show this inequality $x^{n+1}+y^{n+1}\ge x^n+y^n$
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 $$f'(x) = x^{n-1}\cdot \frac{nx^{2n+2} - (n+1)x^{2n+1} + (n+1)x - n}{(x^{n+1} + 1)(x^n + 1)(x^{2n+1} + 1)}.$$
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. $\qquad \Box$