Vtyjkyuk

Let $k$ be a positive integer and $x$ a positive real number. Prove that $(1+x^k)^k+1geq (1+x^k+1)^k$.

This looks similar to Bernoulli's inequality. If we write $X=x^k$, the inequality is equivalent to $(1+X)^frack+1kgeq 1+X^frack+1k$, but this is not exactly in the form where we can apply Bernoulli.

Let $f(X) = (1+X)^r - 1 - X^r$ for $r> 1$. Then $f(0)=0$ and
$$
f'(X) = rleft((1+X)^r-1-X^r-1right)geq 0
$$
since $Xmapsto X^r-1$ is an increasing function. Hence $f$ is also increasing and we get $f(X)geq f(0)=0$ for $Xgeq 0$. 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 $ageq b$ we have $$(a+b,0)succ (a,b),$$ by Karamata we obtain:
$$f(a+b)+f(0)geq f(a)+f(b)$$ or
$$(a+b)^alphageq a^alpha+b^alpha$$ 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 $alpha=frack+1k$ we got your inequality.

Case 1. $xle 1$.

Then $x^kge x^k+1$ 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
$$left(1+left(frac 1x right)^kright)^k+1 ge left(1+left(frac 1x right)^k+1right)^k.$$
This is just Case 1. for the number $dfrac 1x$.

For $rgeq 1$, we have by Minkowski's Inequality that
$$X+1=big(X^r+0^rbig)^frac1r+big(0^r+1^rbig)^frac1rgeq left((X+0)^r+(0+1)^rright)^frac1r=left(1+X^rright)^frac1rtext for all Xgeq0,.$$
This shows that
$$(1+X)^rgeq 1+X^r,.$$
Note that the inequality becomes an equality iff $r=1$ or $X=0$. Now, for $k,linmathbbR_>0$ with $kleq l$, we then see that, with $X:=x^k$ and $r:=fraclk$, we have
$$left(1+x^kright)^fraclkgeq 1+x^l,,text or left(1+x^kright)^lgeqleft(1+x^lright)^k,,$$
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 $f:mathbbR_geq0timesmathbbR_>0tomathbbR$ defined by $$f(x,k):=left(1+x^kright)^frac1ktext for all xgeq0text and k>0$$
is a nondecreasing function in $k$, and it is strictly increasing with respect to $k$ for $x>0$.