# What is the maximum of numbers?

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.

• Well, what result do you expect? – Shaun Sep 28 '17 at 12:54
• Dividing both of the expressions by $a^{1/mn}$, you can eliminate one variable. Might make the analysis easier. Also, raising both expressions to the power of $mn$ means all the $m$ appear in one expression, and all the $n$ appear in the other. Should make the analysis easier. – Arthur Sep 28 '17 at 12:59
• Check Power Means Inequality... – Macavity Sep 28 '17 at 15:41
• I don't want to be pendantic but you mean "larger"; not "largest". I only bring that up because I spent the first five minutes assuming you meant which was the largest possible values for $A$ and $B$ and they are unbounded as $a,b$ can be as large as you like. – fleablood Sep 29 '17 at 16:01

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 $t\in[0,1]$.
• Thanks @Robert Z! With other calculations we have reached an inequality equivalent to $(1+x)^r \geq 1+x^r$ that I wanted to demonstrate without derivatives and that I did not succeed. Is it possible to prove no derivative? – medicu Sep 29 '17 at 15:31