Inequality $(1+x^k)^{k+1}\geq (1+x^{k+1})^k$ Let $k$ be a positive integer and $x$ a positive real number. Prove that $(1+x^k)^{k+1}\geq (1+x^{k+1})^k$.
This looks similar to Bernoulli's inequality. If we write $X=x^k$, the inequality is equivalent to $(1+X)^{\frac{k+1}{k}}\geq 1+X^{\frac{k+1}{k}}$, but this is not exactly in the form where we can apply Bernoulli.
 A: 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 $r = (k+1)/k$. 
A: 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.
A: Case 1. $x\le 1$. 
Then $x^k\ge x^{k+1}$ and therefore
$$(1+x^k)^{k+1} = (1+x^k)(1+x^k)^k \ge (1+x^k)^k \ge (1+x^{k+1})^k.$$
Case 2. $x>1$.
We divide the inequality by $x^{k(k+1)}$ and get the following equivalent form
$$\left(1+\left(\frac 1x \right)^k\right)^{k+1} \ge \left(1+\left(\frac 1x \right)^{k+1}\right)^k.$$
This is just Case 1. for the number $\dfrac 1x$.
A: For $r\geq 1$, we have by Minkowski's Inequality that
$$X+1=\big(X^r+0^r\big)^{\frac{1}{r}}+\big(0^r+1^r\big)^{\frac1r}\geq \left((X+0)^r+(0+1)^r\right)^{\frac{1}{r}}=\left(1+X^r\right)^{\frac{1}{r}}\text{ for all }X\geq0\,.$$
This shows that
$$(1+X)^r\geq 1+X^r\,.$$
Note that the inequality becomes an equality iff $r=1$ or $X=0$.    Now, for $k,l\in\mathbb{R}_{>0}$ with $k\leq l$, we then see that, with $X:=x^k$ and $r:=\frac{l}{k}$, we have
$$\left(1+x^k\right)^{\frac{l}{k}}\geq 1+x^l\,,\text{ or }\left(1+x^k\right)^l\geq\left(1+x^l\right)^k\,,$$
for all $x>0$.  The equality case is when $x=0$ and when $k=l$. This problem is a particular case where $l=k+1$.  Furthermore, we conclude that the function $f:\mathbb{R}_{\geq0}\times\mathbb{R}_{>0}\to\mathbb{R}$ defined by $$f(x,k):=\left(1+x^k\right)^{\frac1k}\text{ for all }x\geq0\text{ and }k>0$$
is a nondecreasing function in $k$, and it is strictly increasing with respect to $k$ for $x>0$.
