L[E(x)] is defined such that it is tangent to the function g at g[E(x)]

Actually it is correct! The proof relies on the convexity inequality, but you also can prove by a tangent plane if you choose L(x)=g(E(x))+g'(E(x))(x-E(x)). thus, as g(x)>=L(x) for all x we got that E(g(x)) >= g(E(x)) for all x.

L[E(X)] and g(E(X)) are both constants, for a random variable X, thus they should be equal to each other

Ryan Tamburrino It is proven in exactly the same manner at the book A First Look at Rigorous Probability Theory by Jeffrey Rosenthal

Yes, you can but as trig functions are periodic, and triangle angles have a condition like at max only one obtuse angle/right angle, you should be very careful as some angles may land on the convex part while others concave!

How did that come about?

it's differentiable at all but finitely many points, so not such a big lie

I will use it anyway since all of u say it is right

bad proof

@@kaidongwei4406 HAHAHHA which part do u not understand

@@itstotallydope his life