If $n\geq m$ then $(x^m+y^m)^{1/m} \ge (x^n+y^n)^{1/n}$

Question

If $n\geq m$ show that: $(x^m+y^m)^{1/m} \ge (x^n+y^n)^{1/n}$, all numbers being positive real.

Obviously since $n\geq m$, for every $a\in \Bbb Z^+$: $a^{1/m}\geq a^{1/n}$, but does it help?

Answer 1

Take the transformation of Michael by $x^m=a$, $y^m=b$ and $\alpha=\frac nm$: $$ x^n=(x^m)^{n/m}=a^\alpha,y^n=(y^m)^{n/m}=b^\alpha. $$ The inequality is equivalent to: $$ ({a}+{b})^{1/m}\geq ({a}^\alpha+{b}^\alpha)^{1/n}\implies $$ $$ ({a}+{b})^{n/m}\geq ({a}^\alpha+{b}^\alpha)\implies $$ $$ (\frac{a}{a+b})^\alpha+(\frac{b}{a+b})^\alpha\leq 1. $$ Now see that for $\alpha>1$: $$ (\frac{a}{a+b})^\alpha\leq \frac{a}{a+b}. $$ and $$ (\frac{b}{a+b})^\alpha\leq \frac{b}{a+b}. $$

Answer 2

Let $x^m=a$, $y^m=b$, $\frac{n}{m}=\alpha$ and $f(x)=x^{\alpha}$.

Thus, $f$ is a convex function because $\alpha\geq1$ and we need to prove that $$(a+b)^{\alpha}\geq a^{\alpha}+b^{\beta},$$ which is Karamata:

Let $a\geq b$.

Thus, since $(a+b,0)\succ(a,b)$, we obtain $$f(a+b)+f(0)\geq f(a)+f(b)$$ and we are done!