Given $a,b,x>0$, $x<y$, prove $(a^x+b^x)^{1/x} > (a^y+b^y)^{1/y}$

Question

I'm thinking about proving $f(x) = (a^x+b^x)^{1/x}$ has negative derivative for all positive $x$.

$$f'(x) = \left(b^x+a^x\right)^\frac{1}{x}\left(\frac{b^x\ln\left(b\right)+a^x\ln\left(a\right)}{\left(b^x+a^x\right)x}-\frac{\ln\left(b^x+a^x\right)}{x^2}\right)$$

To prove this is negative, I need $$x(a^x\ln(a)+b^x\ln(b)) < (a^x+b^x)\ln(a^x+b^x)$$ which is equivalent to

$${a^x}^{a^x}{b^x}^{b^x} < (a^x+b^x)^{a^x+b^x}$$

This looks like a special case of

$$A^AB^B < (A^A+B^B)^{A^A+B^B}$$

for any $A,B>0$. I'm convinced this is true, but I don't know how to prove it either.

Answer 1

Let $a^x=u$, $b^x=v$ and $\frac{y}{x}=\alpha$.

Thus, $\alpha>1$ and we need to prove that $$(u+v)^{\alpha}>u^{\alpha}+v^{\alpha},$$ which is true by Karamata for the convex function $f(x)=x^{\alpha}.$

Indeed, let $u\geq v$.

Thus, $(u+v,0)\succ(u,v)$ and by Karamata $$f(u+v)+f(0)\geq f(u)+f(v),$$ which is our inequality.

A proof of the last inequality without Karamata.

Let $\frac{u}{v}=t$.

Thus, we need to prove that $g(t)>0$, where $$g(t)=(t+1)^{\alpha}-t^{\alpha}-1.$$ Indeed, $$g'(t)=\alpha\left((t+1)^{\alpha-1}-t^{\alpha-1}\right)>0,$$ which says $$g(t)>g(0)=0$$ and we are done again.