For which values of $s>0$ the inequality $x^{s+1}\le (y+z)$ holds true with $1\le x\le y+z$?

Question

Let $s\in\mathbb{R}$, $s>0$. Moreover, let $$1\le x\le y+z.$$

My question is: for which values of $s$ the inequality $$x^{s+1}\le (y+z)$$ holds true?

I see, e.g., that it holds true for $s=1/2$, $x=y=2, z=1$, but I just can't to find a general relation between $p, s$ in order to make it true in general.

What I can do more than write

$$x\le (y+z)^{1/(s+1)}?$$

Could someone please help?

Thank you in advance?

Answer 1

For $x=1$ the inequality always holds. For $x>1$ we have \begin{align} x^{s+1} \le w & \iff (s+1)\log x\le \log w\\ &\iff s+1 \le \frac{\log w}{\log x}\\ &\iff s \le \frac{\log w}{\log x}-1 \end{align}