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**running-gag** Hi

The inequality is true for $\displaystyle x \geq 0$

For instance for k=2 : $\displaystyle x^3-3x+2 \leq 0$ for x=-3

There are several ways to show the inequality

You can study $\displaystyle f(x) = x^{k+1}-(k+1)x+k$

f is differentiable and $\displaystyle f'(x) = (k+1)(x^k-1)$

The minimum of f is obtained for x=1 and f(1)=0 therefore $\displaystyle f(x) \geq 0$

Second way : since f(1)=0 and f'(1)=0, f(x) can be factorized by (x-1)²

You can show that $\displaystyle f(x) = (x-1) \left(\sum_{j=1}^{k}x^j-k\right)$

and $\displaystyle f(x) = (x-1)^2 \left(\sum_{j=0}^{k-1}(k-j)x^j\right)$

Both factors are positive for x positive