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.

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 $\dfrac 1x$.
