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
 A: 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
$$f'(x) = x^{n-1}\cdot \frac{nx^{2n+2} - (n+1)x^{2n+1} + (n+1)x - n}{(x^{n+1} + 1)(x^n + 1)(x^{2n+1} + 1)}.$$
A: 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. $\qquad \Box$
A: A different approach is to consider the equation $x^{2n+1}+y^{2n+1}=k$ and take polar coordinates. This gives $(x,y)=(k^a\cos^{2a}\varphi,k^a\sin^{2a}\varphi)$ where $a=1/(2n+1)$. We wish to show that* if $k\ge2$ then $$\frac{x^{n+1}+y^{n+1}}{x^n+y^n}=k^a\cdot\frac{\cos^{1+a}\varphi+\sin^{1+a}\varphi}{\cos^{1-a}\varphi+\sin^{1-a}\varphi}\ge1$$ for all $a\in(0,1)$ which we can rewrite as $$k^a\ge\sec^{2a}\varphi\cdot\frac{1+\tan^{1-a}\varphi}{1+\tan^{1+a}\varphi}=\frac{(1+s^2)^a(1+s^{1-a})}{1+s^{1+a}}$$ where $s=\tan\varphi\in[0,1]$ as the function is invariant under the transformation $s\mapsto1/s$. This means that $$k^{1-2b}\ge\frac{(1+t)^{1-2b}(1+t^a)}{(1+t^{1-a})}\implies\left(\frac{1+t}k\right)^{1-2b}-\frac{1+t^{1-b}}{1+t^b}\le0$$ where $t\in[0,1]$ and $b=(1-a)/2\in(0,1/2)$. Since the LHS is a decreasing function of $k$ it remains to show that $$\max_t\left\{\left(\frac{1+t}2\right)^{1-2b}-\frac{1+t^{1-b}}{1+t^b}\right\}=0.$$ The easiest way to prove this uses calculus and I don't think there are methods based solely on inequality results to do so.
 *This is actually a stronger claim, that for all real $x,y,n>0$, if $x^{2n+1}+y^{2n+1}\ge2$ then $x^{n+1}+y^{n+1}\ge x^n+y^n$. 
