Inequality $(1+x^k)^{k+1}\geq (1+x^{k+1})^k$

Let $f(X) = (1+X)^{r} - 1 - X^{r}$ for $r> 1$. Then $f(0)=0$ and $$ f'(X) = r\left((1+X)^{r-1}-X^{r-1}\right)\geq 0 $$ since $X\mapsto X^{r-1}$ is an increasing function. Hence $f$ is also increasing and we get $f(X)\geq f(0)=0$ for $X\geq 0$. Now set $r = (k+1)/k$.

Let $f(x)=x^{\alpha}$, where $\alpha>1$.

Thus, $f$ is a convex function and since for positives $a$ and $b$ such that $a\geq b$ we have $$(a+b,0)\succ (a,b),$$ by Karamata we obtain: $$f(a+b)+f(0)\geq f(a)+f(b)$$ or $$(a+b)^{\alpha}\geq a^{\alpha}+b^{\alpha}$$ and since the last inequality is symmetric, it's true for all positives $a$ and $b$.

Id est, for $a=1$, $b=x^{k}$ and $\alpha=\frac{k+1}{k}$ we got your inequality.