which is larger??

Рет қаралды 69,098

3 жыл бұрын

We look at a nice inequality.
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Пікірлер: 172

@Carhill 3 жыл бұрын

9:50 - that minus sign added in post scared me then... thought a mass array of pixels had died on my monitor.

@badreddinebencherki6729 3 жыл бұрын

me too xD

@JoeCMath 3 жыл бұрын

Ah yes, the classic sweatshirt minus sign summon, a rare find!

@graysonking16 2 жыл бұрын

Pre-watch guess: a^a + b^b is larger because either a or b is larger, and exponentials grow very quickly, so being able to raise the larger number to itself probably makes up for the fact that you would get to raise the smaller number to the larger number in the opposing case.

@kajamix 3 жыл бұрын

Much simpler proof: We want to prove a^b + b^a

@Milan_Openfeint 3 жыл бұрын

just filling the missing steps... a^(a+x) + (a+x)^a

@marioytambor 3 жыл бұрын

That's pretty neat

@arremangamelpicoapeos9853 3 жыл бұрын

Very nice

@jebooiii6462 3 жыл бұрын

Came here to say this but you beat me to it :p

@speedsterh 3 жыл бұрын

Excellent find !

@ShinySwalot 3 жыл бұрын

It's very intuitive too, either you're doing the highest value to the highest power or the highest value to the lowest power.

@vinc17fr 3 жыл бұрын

I was wondering something like that and thinking about a generalization: Let 0 < a ≤ b and 0 < u ≤ v. Which is larger a^u + b^v or a^v + b^u? I suppose that it is a^u + b^v, but I haven't tried to prove it.

@ShinySwalot 3 жыл бұрын

@@vinc17fr Hmmm, that's actually a really interesting idea, I wonder if that's enough restriction in order to generalize it though.

@vinc17fr 3 жыл бұрын

@@ShinySwalot Actually, that's not true, at least for b < 1. Here's a counter-example: (1/3)^2 + (1/2)^1000 < (1/3)^1000 + (1/2)^2.

@vinc17fr 3 жыл бұрын

@@ShinySwalot I think that one needs to consider the different cases, something like whether the values are less or greater than 1. Using u = a and v = b like in the original problem makes these conditions more restrictive.

@ShinySwalot 3 жыл бұрын

@@vinc17fr Or just make it over the natural numbers instead? Just proposing something lol

@knvcsg1839 3 жыл бұрын

I love the way you deduce the approach from fine details of the problem. Loved it.

@Yash-ql7vn 3 жыл бұрын

Hey, just discovered your channel through recommendations, thanks for making a great video. Just subscribed! Hoping for more quality content 💛

@OuroborosVengeance 3 жыл бұрын

Its a goldmine

@marcrenard515 3 жыл бұрын

This two quantities are rate increase of f:x->ln(x), they represant the slope of secant to the curve. Since ln(x) is concave ( f''(x) =-1/x²(ln(x)-ln(a)) /(x-a) is decreasing. We have b

@andimandi8491 3 жыл бұрын

I had the conclusion that a^a+b^b=>a^b+b^a because there are always 3 cases: case 1. a=b; case 2. a>b; case 3. ab^a+a^b.

@mathunt1130 3 жыл бұрын

As he said, wlog b>a, then fix a. Examine the function f(b)=a^a+b^b-a^b-b^a, and compute f'(b) and examine the sign. It's late at night and this was the first thing I thought of.

@BilalAhmed-wo6fe Жыл бұрын

Did you solve it ?

@manjunathbhat6934 3 жыл бұрын

A simpler approach : Let's scale down a and b wrto a. Let a = 1, b /a = t, t is a positive real number. Hence X = a^b + b^a = 1 + t, and Y = a^a + b^b = 1 + t^t. Compare X and Y, implies comparing t and t^t, implies comparing log(t) and t*log(t). t^t > t for all positive real t, as log(t) * (t - 1) > 0 for all positive real t. Hence Y > X.

@jebooiii6462 3 жыл бұрын

That's really nice

@vinc17fr 3 жыл бұрын

But how do you prove that you can scale down a and b? Since they appear also in exponent, this is not obvious.

@Notthatkindofdr 3 жыл бұрын

@@vinc17fr Agreed. There is no way to "scale" the problem to look like this. This solution really only considers the case that a=1.

@A_Helder16 3 жыл бұрын

Love Your Algebra Series Videos... Can you make a video review of the three parts of Sylow's Theorem

@marioytambor 3 жыл бұрын

On the last part, could we just divide by b-a, since it's positive but bigger than 0? I feel then the homework part is easier ( because adding log(t)/(b-a) and differentiating)

@bryguy6651 3 жыл бұрын

This is very similar to the Rearrangement Inequality!

@mehmeterciyas6844 3 жыл бұрын

Hi from future!

@arnabchatterjee4847 3 жыл бұрын

@@mehmeterciyas6844 Hi I am from past

@sirlight4954 3 жыл бұрын

Indeed it is. There is a generalisation of this inequality for arbitrary functions -- if we have x1

@tomatrix7525 3 жыл бұрын

The proof of the final one is quite simple. Here is a rough outline. Given 0

@keithmasumoto9698 3 жыл бұрын

Proof by contradiction: Say -lna/(1-a) > (lnb-lna)/(b-a). Then doing some cross multiplication, we arrive at lna/(1-a)>lnb/(1-b) which is not true for a

@megauser8512 3 жыл бұрын

Yes, you are correct, since: -lna / (1-a) > (lnb-lna) / (b-a) implies that -lna * (b-a) > (lnb-lna) * (1-a), which leads to -b*lna+a*lna > lnb-lna-a*lnb+a*lna, so -b*lna > lnb-lna-a*lnb, so lna-b*lna > lnb-a*lnb, so lna * (1-b) > lnb * (1-a), so lna/(1-a) > lnb/(1-b), but since 0 < a < b < 1, then -1 < -b < -a < 0 and lna < lnb < 0, so 0 < -lnb < -lna, so we also have: 0 < 1-b < 1-a < 1, so 1 < 1/(1-a) < 1/(1-b), so -1/(1-b) < -1/(1-a) < -1, so -lnb/(1-b) < -lnb/(1-a) < -lna/(1-a), so lna/(1-a) < lnb/(1-b), which is a contradiction with the above.

@timurpryadilin8830 3 жыл бұрын

Homework: let f(t) = (ln t- ln a) / (t-a). Then f'(t) = ((t-a)/t-lnt+lna) / (t-a)^2 has the same sign as 1-a/t + ln(a/t). Now let g(y) =1-y+lny, so that f'(t) = g(a/t) . g'(y) = 1/y-1 > 0 when ya. g(1)=0, so g(y) < 0 for y

@karangupta1825 3 жыл бұрын

Try this : ((a(subi))^2)) + ((a(subi))+1)^2 = ((b(subi))^2) & ((x(subi))^2) -1 = 2((y(subi))^2) Edit: Then prove: (x(subi))+(y(subi)) = (b(sub(i+1))) (x(subi))-(y(subi)) = (b(sub(i)))

@kongkong5 3 жыл бұрын

If you let f(t) = ln t, much easier

@johnnath4137 3 жыл бұрын

(ln1 - lna)/(1 - a) ≤ (lnb - lna)/(b - a) ↔ -blna + alna ≤ lnb - lna - alnb + alna ↔ -blna ≤ lnb - lna - alnb ↔ ln(ba^b/ab^a) ≥ 0. But (ba^b/ab^a) ≥ (ba^a/ab^a) = (b/a)^(1 - a) ≥ 1 → ln(ba^b/ab^a) ≥ ln(ba^a/ab^a) ≥ 0 → (ln1 - lna)/(1 - a) ≤ (lnb - lna)/(b - a).

@dpeastman 3 жыл бұрын

Here's my much simpler solution: Assume a

@Nyarmith 3 жыл бұрын

wow, these videos are actually fun

@lawlietl1370 3 жыл бұрын

Is possible obtain a factorization for (a^b) +/- (b^a) ?

@romajimamulo 3 жыл бұрын

Homework: cross multiply (which doesn't flip the inequality because the denominators are positive), then algebraically rearrange to (ln(a))/(1-a) ≤ (ln(b))/(1-b) Then, notice how this is like "prove that f(x)=(ln(x))/(1-x) is increasing on the range (0,1)" Which can then be done by taking the derivative and checking for critical points

@fix5072 3 жыл бұрын

Nicely done! trying to check for the critical point is not easy with trying to apply algebra, since you'd have to use Lambert W function (x=-W(-1/e)=1), but trying nice values will eventually give you the answer that x>0 except for x=1 which is however not defined for the function ln(x)/(1-x).

@oscarengland6122 3 жыл бұрын

To prove that f(x)=(ln(x))/(1-x) is increasing would it be enough to observe that the numerator is increasing and the denominator is decreasing therefore f(x) must be increasing?

@romajimamulo 3 жыл бұрын

@@oscarengland6122 Yes, on the range (0,1) that does work

@maxwellsequation4887 3 жыл бұрын

How is there comments from 3 weeks ago when the video is just out

@fix5072 3 жыл бұрын

channel members have early access to the videos

@MichaelPennMath 3 жыл бұрын

This is not true! I upload ahead of time and leave videos unlisted. I am not sure how people find them...

@fix5072 3 жыл бұрын

@@MichaelPennMath Well I'm sorry then. As many KZfaqrs do so I thought it woulb be the same with You since its a nice way to reward channel members.

@goodplacetostop2973 3 жыл бұрын

@@MichaelPennMath Here's the trick... Go on playlists page, sort them by last video added and there we go. In some playlists, there are some unlisted videos but NOT private which means you can watch them even if they're not officially published.

@abhisheknanda85 3 жыл бұрын

@@MichaelPennMath Adding an unlisted video into a public playlist effectively makes the video public too.

@charlottedarroch 3 жыл бұрын

This inequality reminded me of the rearrangement inequality and in fact a generisation of this inequality is very similar to the rearrangment inequality. The inequality in the video can be extended to the result: If x_1 >= x_2 >= ... >= x_n > 0 and y_1 >= y_2 >= ... >= y_n > 0, then x_1^y_1+x_2^y_2+...+x_n^y_n >= x_1^y_p(1)+x_2^y_p(2)+...+x_n^y_p(n) for any permutation p on the set {1,2,...,n}. This is proved by induction using exactly the same argument as is used to prove the rearrangement inequality, starting from the base case that if x_1 >= x_2 > 0 and y_1 >= y_2 > 0, then x_1^y_1+x_2^y^2 >= x_1^y_2+x_2^y_1. The base case can be proved by first observing the following equivalences: x_1^y_1+x_2^y^2 >= x_1^y_2+x_2^y_1 iff x_1^y_1-x_1^y_2 >= x_2^y_1-x_2^y^2 iff e^(y_1ln(x_1))-e^(y_2ln(x_1)) >= e^(y_1ln(x_2))-e^(y_2ln(x_2)) iff (e^((y_1-y_2)ln(x_1))-1)e^(y_2ln(x_1)) >= (e^((y_1-y_2)ln(x_2))-1)e^(y_2ln(x_2)). As the logarithm and exponential are increasing and since y_2 is positive: x_1 >= x_2 implies ln(x_1) >= ln(x_2) implies y_2ln(x_1) >= y_2ln(x_2) implies e^(y_2ln(x_1)) >= e^(y_2ln(x_2)). Then for the other terms, since y_1 >= y_2, we have y_1-y_2 >= 0, so: ln(x_1) >= ln(x_2) implies (y_1-y_2)ln(x_1) >= (y_1-y_2)ln(x_2) implies e^((y_1-y_2)ln(x_1))-1 >= e^((y_1-y_2)ln(x_2))-1 Finally, ince e^(y_2ln(x_1)) >= e^(y_2ln(x_2)) > 0, we may multiply the two inequalities to obtain the result.

@Notthatkindofdr 3 жыл бұрын

Unfortunately this proof has the same flaw as others in the comments. The two sides of your inequality e^((y_1-y_2)ln(x_1))-1 >= e^((y_1-y_2)ln(x_2))-1 (this would be easier to read as (x_1)^(y_1-y_2)-1 >= (x_2)^(y_1-y_2)-1, by the way) could both be negative, and then you cannot multiply it by the other inequality. (To see what I mean, note that -2 > -3 and 5 > 3, but -10 < -9.) This case corresponds to the second "branch" that Michael does in the video.

@Notthatkindofdr 3 жыл бұрын

In fact, @Vincent Lefèvre has a counterexample to your generalization: (1/3)^2 + (1/2)^1000 < (1/3)^1000 + (1/2)^2 in the comments below.

@charlottedarroch 3 жыл бұрын

@@Notthatkindofdr Oh yes, you're right. If we change the constraint to x_1 >= x_2 >= ... >= x_n >= 1, then the result holds. But one can certainly do better. Though the details will be more complicated than the rearrangement inequality.

@user-qp6kd9hf9g 3 жыл бұрын

Brilliant video

@dneary 3 жыл бұрын

I split out the equality, and followed the LHS method myself with a

@replicaacliper 3 жыл бұрын

Why does that only work if b-a>=1? x^k is strictly increasing on x>=0 whenever k>0, then a^(b-a)-1=1, then both of these quantities are positive, and since 0

@dneary 3 жыл бұрын

@@replicaacliper I need to look back, but... a^b+b^a < a^a+b^b \implies a^b-a^a < b^b-b^a \implies a^a(a^(b-a)-1) < b^a(b^(b-a)-1) and since we assumed 0

@TheSmilodon2 3 жыл бұрын

i took a^b + b^a < b^b + a^a i noted c = b-a (with presumption b > a ) we have: a^a*a^c + b^a < b^a*b^c + a^a => a^a*(a^c-1) < b^a*(b^c-1) => (a/b)^a < (b^c-1)/(a^c-1) since a

@HideyukiWatanabe 3 жыл бұрын

It holds that a^a + b^b >= 2((a+b)/2)^((a+b)/2) >= a^b + b^a. Use downward convexity of f(x)=x^x and Jensen's inequality implies the former. The latter needs some messy calculus.

@MrMetrizable 3 жыл бұрын

-ln x is convex

@tanimahmed8095 3 жыл бұрын

@Michael Penn sir, your videos are awesome. Can you please make a video on the proof of L' Hospital's Rule?

@craftexx15 3 жыл бұрын

He already did. Look up in his real analysis course playlist.

@jakobhablitz1429 3 жыл бұрын

I burnt my poptart after getting distracted watching this

@prithujsarkar2010 3 жыл бұрын

3 weeks ago commented ? huh ?

@Tiqerboy 3 жыл бұрын

If you switched to Kellogg's Frosted Flakes they would have gone soggy in milk. Can't win, even with a tiger on your side.

@jakobhablitz1429 3 жыл бұрын

@@prithujsarkar2010 I'm simply built different

@prithujsarkar2010 3 жыл бұрын

@@jakobhablitz1429 lmao

@FisicoNuclearCuantico 3 жыл бұрын

@Jakob Hablitz 🙁

@FisicoNuclearCuantico 3 жыл бұрын

Best regards Mike.

@dacianbonta2840 3 жыл бұрын

Chapter 14 of Barnard and Child "Higher Algebra" for a calculus-free approach to this topic. Also, the post of kajamix

@nopenoper1980 3 жыл бұрын

Given b>a>0 a^b-a^a

@AAA-mv7dv 2 жыл бұрын

Given that a,b are positive real numbers, further suppose that a>b and try to find such values for the inequality a^a + b^b < a^b + b^a and you'll arrive at the conclusion that it's never true for such values.

@inquirusinq6060 3 жыл бұрын

At 2:42 you write an unequality a^b-a^a = a^b-a^a if b>=a. Sry for eng. Not my native ;3

@rayansoufargi1439 2 жыл бұрын

Ln(x)-ln(a)/x-a is obviously decreasing and b

@annanemustapha7279 3 жыл бұрын

Witch is larger ? ( aⁿ + b) or (a + bⁿ) if a

@nightmare9566 2 жыл бұрын

a^a + b^b because it ensures that the largest number is raised to the largest exponent. either a is larger and then you have a^a, or b is larger and you get b^b. Not exactly rigorous, but do we really need more than that immediate intuition? 3^4 + 4^3 < 3^3 + 4^4, 100^2 + 2^100 < 100^100 + 2^2 etc.

@sevdanaamani3266 3 жыл бұрын

Convex function and its done.QED

@purity_one 3 жыл бұрын

Wait, we have proved it only for b≥1 and b∈(0,1) (considering "homework"). But what about negative values of b? (cuz in the task we suppose b is Rational) Or is there something I didn't understand?

@davode76166 3 жыл бұрын

Consider R+ in the first line of the problem

@Paulo-jh8vz 3 жыл бұрын

b it's a positive real number

@purity_one 3 жыл бұрын

Oh, R+ means real positive numbers. Got it, thx

@no_one6749 3 жыл бұрын

explaining >= doing

@MrKinyodude 3 жыл бұрын

Michael, you are right about skin tones ... white balance is quite off, whole scene is reddish. If camera cannot compensate, consider changing light source. Great math as usual!

@sarthakkumarbehera1015 3 жыл бұрын

Hi from future

@laurensiusfabianussteven6518 3 жыл бұрын

I thought there will be mean-value theorem...

@mtaur4113 3 жыл бұрын

There might be later...

@sevdanaamani3266 3 жыл бұрын

Y so mean?

@josephcheng5949 3 жыл бұрын

3:33 that escalated quickly from algebra to calculus! 😱 😂

@mtaur4113 3 жыл бұрын

If we only look at whole numbers, some other argument might be possible, but it might not be easy without calculus... If we allow non-whole numbers, it was always calculus all along... :)

@MrRyanroberson1 3 жыл бұрын

What if... they're negative numners from the set x/(2y+1) for x,y integers?

@MrRyanroberson1 3 жыл бұрын

And finally what if one is negative?

@hq4889 3 жыл бұрын

Hii Michael please solve this problem 👇 Find +ve integers a and b such that (a^1/3 +b^1/3 -1) ^2 = 49 +20 (6) ^1/3

@ricardocavalcanti3343 3 жыл бұрын

{a, b} = {48, 288}

@3snoW_ 3 жыл бұрын

5:06 How is this not always true? if 0 < a < x < b then both ln(a) < ln(b) and a^x < b^x are true, hence their product must also share the same relationship: ln(a) * a^x < ln(b) * b^x regardless of whether b is greater than 1 or not

@subinmdr 3 жыл бұрын

I think its because ln a and ln b are negative when b

@einsteingonzalez4336 3 жыл бұрын

I thought that the problem is similar to a^b > b^a, but this isn't. Whoops!

@sirlight4954 3 жыл бұрын

This reminded me of a rearrangement inequality, which stated that for all x1,y1,x2,y2>0 x1y1+x2y2>x1y2+x2y1. This generalises to arbitrary number of summands. And what is surprising is that this inequality generalises even to arbitrary functions. If we have x1

@Notthatkindofdr 3 жыл бұрын

Unfortunately the inequality f1'(x)f2'(1/2).

@sirlight4954 3 жыл бұрын

@@Notthatkindofdr Ah, yes, sorry, we need a^x, not x^a as a function. Then the derivative will be ln a * a^x which is always bigger than ln b * b^x if a>b

@Notthatkindofdr 3 жыл бұрын

@@sirlight4954 Sorry, I don't think that is quite true either. Try a=e^(-.95), b=e^(-6), x=e^(-1), for example. Then a>x>b, but ln(a)a^x=-0.67

@sirlight4954 3 жыл бұрын

@@Notthatkindofdr yeah, sadly it looks like this approach only works for a,b>1

@junkdubious Жыл бұрын

Wonder if this easier if you evaluate these as limits.

@mikegallegos7 3 жыл бұрын

HOMEWORK !!! YAAAAaaaaahhggggghh !! LOL

@karangupta1825 3 жыл бұрын

Sir, please make some videos on continued fractions, infinite series and telescopic series.

@karangupta1825 3 жыл бұрын

@@guydror7297 you are saying that to whom To me Or To the YouThoober

@piojo1987 3 жыл бұрын

The last step on the board can be proved by seeing that the ln function has negative second derivative. The expression (ln(x)-ln(a)/(x-a)) is the avarage of the slope of the ln function between a and x, and because ln has negative second derivative, increasing x with fixed a decreases the result of that average. As x=b is lower than x=1, the average slope from a to b is greater than the average slope of ln from a to 1, the HW inequality is proved. And that’s a good place to stop!

@andrewmcnamara3959 3 жыл бұрын

Got lost fairly quickly on this one

@Giyga 3 жыл бұрын

My first intuition was 1^1+2^2 or 1^2+2^1

@gourabjitbiswas 3 жыл бұрын

🔥🔥🔥

@ollllj 2 жыл бұрын

complex numbers feel left out of all the fun inequality. maybe inequality is just incomplete (and not all that useful)

@fatalaf 3 жыл бұрын

Those are some long hoodie strings my guy

@zhuolovesmath7483 3 жыл бұрын

let f(x)=(lnx-lna)/(x-a) All we need to prove is that f'(x)1+lnx then we'll get the answer

@mtaur4113 3 жыл бұрын

Let a belong to the interval (0,1). Inequality at 10:25 is true for all x in (a,b) if and only if it is true for x = a, so let x = a. Let g(t) = [ln(t) - ln(a)]/[t - a] We want to prove that -ln(a) < g(t) for all t in (a, 1). t=1 is shown at 12:08. If we can show that g'(t) g(1) > -ln(a) for all such t. g'(t) can be done by the quotient rule. To solve the inequality g'(t)>0, compute g' and multiply by the denominator, (x-a)^2. Simplify to get: 1/t < [ln(t) - ln(a)]/(t - a) The right-hand side is the average rate of change of ln on the interval (a,t). By the mean value theorem, that is equal to the derivative of ln at some intermediate point, that is, "ln-prime of c" or 1/c, where c is in (a,t) So we want to solve for all t such that 1/t < 1/c. This is in fact all t in (a,1), because c < t and both numbers are positive.

@ayoubshams261 2 жыл бұрын

thanks for btfl problems and sltns

@donl.3731 3 жыл бұрын

all im hearing is "m to the b, m to the b, m m m m m to the b"

@daniello4038 3 жыл бұрын

The inequality can also be proved by letting b=ax, x>=1.

@Notthatkindofdr 3 жыл бұрын

How?

@toddbiesel4288 3 жыл бұрын

So a^b + b^a is never larger? Mamma mia!

@josephcheng5949 3 жыл бұрын

OMG @Michael Penn you give homework now?! 😆

@prathmeshraut1616 3 жыл бұрын

Need a like from Micheal penn

@alimoharam4362 2 жыл бұрын

this problem reminds me of a song

@williammauriciogiraldomuri9855 3 жыл бұрын

I think proof for the homework might go like this: - lnx

@nekrozvalkyrus1754 3 жыл бұрын

Does this not work? For a=b they are equal Suppose b>a (b/a)^b *a^b +b^a=b^b +b^a (b/a)^a *a^a +b^b=b^b+b^a ((b/a)^b -1)a^b +a^b +b^a=b^a +b^b=((b/a)^a -1)a^a +a^a +b^b ((b/a)^b -1)a^b>((b/a)^a -1)a^a Therefore a^a+b^b>=a^b +b^a Edit: I didn't consider the case of a

@goodplacetostop2973 3 жыл бұрын

12:13 Homework 12:37 Good place to stop

@vaxjoaberg9452 3 жыл бұрын

12:37 Good place to sto

@yeahyeah54 3 жыл бұрын

obviously a^a +b^b is bigger. this is my thought: i took a=1,b=2 and that's all folks

@yeahyeah54 3 жыл бұрын

lol Michael did the same thing, i didn't see the video when i was writing

@leefisher6366 3 жыл бұрын

Well, two squared is four, three cubed is twenty-seven, and that makes thirty one. Alternatively, two cubed is eight, and three squared is nine, only giving seventeen. In general, if a

@jesusalej1 3 жыл бұрын

Who are those kids? 5:50

@mattcassar4749 3 жыл бұрын

Edit: there is a mistake, so this proof only works for b >= 1. I'm not going to extend it to handle the a, b in (0, 1) case, but it shouldn't be too hard to fix it :) I think there's a simpler proof, but I might have made a mistake. WLOG assume a

@Notthatkindofdr 3 жыл бұрын

I had thought of an algebraic solution like this too, but as you say there is flaw when b= a^a(1 - a^(b - a)) + a^a(b^(b - a) - 1) does not follow if b^(b-a)-1

@TheUltiCAKE 3 жыл бұрын

Put any number into a and b (as long as they are not the same and not 0) and find out...ez pz

@wepped482 3 жыл бұрын

Yeah, only if you are talking about positive numbers.. which is your suppose statement. Okays.

@sabitasahoo5388 3 жыл бұрын

One of the most underrated channel in KZfaq.

@speedsterh 3 жыл бұрын

I love it too, but the entry level math-wise is pretty steep for someone who hasn't studied maths at college

@random-td7tf 3 жыл бұрын

Nice, I see time travellers here.

@donaldbiden7927 3 жыл бұрын

😂 😆

@goodplacetostop2973 3 жыл бұрын

🙈

@random-td7tf 3 жыл бұрын

@@goodplacetostop2973 How did you find the unlisted video ?

@AcaciaAvenue 3 жыл бұрын

I solved it in a different way, please tell me if I did any mistake: I need to prove that a^a + b^b >= a^b + b^a. Suppose than a>b. I can then write a = b+c, with c>0. I replace this in the above inequality: (b+c)^(b+c) + b^b >= (b+c)^b + b^(b+c) I bring the (b+c) terms on the left and the rest on the right: (b+c)^(b+c) - (b+c)^b >= b^(b+c) - b^b I then split the terms with (b+c) on the exponent: [(b+c)^b]*[(b+c)^c] - (b+c)^b >= (b^b)*(b^c) - b^b I factor (b+c)^b on the left side and b^b on the right side [(b+c)^b] * [(b+c)^c - 1] >= b^b * (b^c - 1) I divide bot terms for b^b and "merge" the first term [(1+c/b)^b] * [(b+c)^c - 1] >= b^c - 1 Since c>0 and b>0, c/b is >0 as well, so 1+c/b is greater than 1. Any number greater than 1 powered by a number greater than 0 gives a number greater than 1, so (1+c/b)^b is greater than 1. Also, since c>0, (b+c)^c is surely greater than b^c, and that stays true if I subtract 1 from both. So on the left side I have something bigger than the right side multiplied by something bigger than 1, so the inequality is surely true.

@keinKlarname 3 жыл бұрын

@Hekatos: Great solution!

@Notthatkindofdr 3 жыл бұрын

Your last step has a flaw, unfortunately. If a(b+c)^c-1>b^c-1, and if both sides of the inequality are negative then you cannot just multiply it by the inequality (1+c/b)^b>1. (For example, -2>-3 and 2>1, but -4

@hsjkdsgd 3 жыл бұрын

Definitely a hard problem to figure out

@TimesOfSilence 3 жыл бұрын

Let's make a game and count how many times Michael Penn uses for the word "but" in the beginning of a sentence that doesn''t go in a different direction than the sentence before. Example: "1+2+3 is equal to 3+3. But that is equal to 6." My guess without having seen the whole video yet is 40 times.