<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] Since at the point $x=y$, equality holds, and  multiple derivatives are $0$, it appears likely that a solution with derivatives is needed. Here is one:
Following @River Li's calculations, it suffices to prove, for $x \ge 1$,  that (let $n$ be a parameter)
$$
f(x) = \ln (x^{n+1} + 1) - \ln (x^n + 1)- \frac{1}{2n+1}\ln \frac{x^{2n+1} + 1}{2} \ge 0
$$
This holds with equality for $x=1$, so we prove that $f(x)$ is increasing with $x$. The derivative is
$$
f'(x) = \frac1x \Big[ \frac{(n+1)x^{n+1}}{x^{n+1} + 1} - \frac{nx^{n}}{x^{n} + 1}- \frac{x^{2n +1}}{x^{2n +1} + 1} \Big]
$$
Multiplying out gives
$$
f'(x) = \frac{x^{n-1}}{ (x^{n+1} + 1) (x^{n} + 1) (x^{2n +1} + 1)} \Big[x - n + nx - x^{2n +1} + n x^{2n +2} - n x^{2n +1}\Big]
$$
Hence it suffices to prove that the last term in brackets is $\ge 0$, slightly rewritten:
$$
g(x) = x(n+1) - n  + x^{2n +1}(-1 -n + n x)\ge 0
$$
Again we have equality for $x=1$. Further, note that $x(n+1) - n \ge 0$ holds always for $x \ge 1$, and $-1 -n + n x \ge 0 $ holds for $x\ge \frac{n+1}{n}$, so we only need to consider the range $x \in [1 \; \frac{n+1}{n}]$. In that range, we need to prove
$$
 \frac{x(n+1) - n  }{1 + n - n x} \ge x^{2n +1} 
$$
Taking logarithms gives
$$
h(x) =  \ln (x(n+1) - n  ) - \ln (1 + n - n x) - (2n + 1 ) \ln x \ge 0 
$$
Since we have equality for $x=1$, we use the above method again and prove that $h(x)$ is increasing. Taking derivatives gives that we need to prove
$$
h'(x) =   \frac{n+1}{x(n+1) - n  }+  \frac{n}{1 + n - n x} - (2n + 1 ) \frac1x \ge 0
$$
or, multiplying out,
$$
h'(x) =   \frac{1}{x(x(n+1) - n)  (1 + n - n x)} \Bigg[ n(2n^2 + 3n + 1)(x - 1)^2 \Bigg] \ge 0
$$
But this is clearly true for $x \in [1 \; \frac{n+1}{n}]$, which proves the claim. $\qquad \Box$
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