<s>[INST] show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$ let $x,y>0$ and $n$ be positive integer.if
$$x^{2n+1}+y^{2n+1}\ge 2$$
show that
$$x^{n+1}+y^{n+1}\ge x^n+y^n$$
maybe use Holder inequality: for example
$$(x^{n+1}+y^{n+1})^n(1+1)\ge (x^n+y^n)^{n+1}$$
so we must prove
$$\dfrac{1}{2}(x^n+y^n)^{n+1}\ge (x^n+y^n)^n$$
or
$$x^n+y^n\ge 2$$
this last inequality seem it is not hold.so How to prove my question.Thanks
 [/INST] Hint:
It suffices to prove that
$$\frac{x^{n+1} + y^{n+1}}{x^n + y^n} \ge \left(\frac{x^{2n+1} + y^{2n+1}}{2}\right)^{\frac{1}{2n+1}}.$$
Since this inequality is symmetric and homogeneous, assume $y = 1$ and $x \ge 1$. It suffices to prove that
$$\ln (x^{n+1} + 1) - \ln (x^n + 1) \ge \frac{1}{2n+1}\ln \frac{x^{2n+1} + 1}{2}.$$
Take derivative.
Update
Let $f(x) = \mathrm{LHS} - \mathrm{RHS}$. We have
$$f'(x) = x^{n-1}\cdot \frac{nx^{2n+2} - (n+1)x^{2n+1} + (n+1)x - n}{(x^{n+1} + 1)(x^n + 1)(x^{2n+1} + 1)}.$$
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