Proof in number-theory

Question

Is there exist any proof for this statement? $a^x + b^x > \left(\frac{a + b}{2}\right)^x$ for any $x > 1$. I am interested in values between $0$ and $1$ for $a$ and $b$. I appreciate it if you provide a general proof (e.g. series) for this statement.

Thanks

Answer 1

As @dxiv notes in the comments, a stronger inequality is true.

First observe that $f(t) = t^x$ is a convex function for $x > 1$, $t>0$ which you can check with calculus. Then, $f(\frac{a+b}{2}) \leq \frac{1}{2} (f(a) + f(b))$, giving the required result.

Answer 2

Let $a>b$

Your equation results in $$a^x+b^x=S^x$$

Where $$b<S<a$$

Thus $$b^x<S^x<a^x$$

Seeing the original equation using this, its clear the LHS is far greater than the RHS.

Because $$a^x>S^x; b^x>0$$