TITLE: What is the maximum of numbers?
QUESTION [6 upvotes]: Let $a, b \in (1, \infty)$ and $ m,n $ natural numbers at least equal to $2$ with $a\leq b$ and $m\leq n$.
Which is the largest of the numbers 
$$ A =(a^{\frac{1}{n}}+b^{\frac{1}{n}})^{\frac{1}{m}}$$ and $$ B=(a^{\frac{1}{m}}+b^{\frac{1}{m}})^{\frac{1}{n}} ?$$
We applied the inequality of generalized averages, calculations with radicals that did not lead to the expected response.

REPLY [5 votes]: Note that $A\geq B$ iff
$$a^{1/m}(1+x)^r=(a^{1/n}+b^{1/n})^{n/m}=A^{n}\geq B^{n}=(a^{1/m}+b^{1/m})=a^{1/m}(1+x^r)$$
where $x=(b/a)^{1/n}\geq 1$, and $r=n/m\geq 1$.
So it remains to show that $(1+x)^r\geq (1+x^r)$, that is
$$\left(\frac{1}{1+x}\right)^r + \left( \frac{x}{1+x}\right)^r\leq 1$$
which holds because $t^r\leq t$ for $t\in[0,1]$.