TITLE: Show that $2^{\sqrt{2}}>1+\sqrt{2}$
QUESTION [5 upvotes]: Given that $\sqrt{2}>1.4$ and $(1+\sqrt{2})^5<99$, I need to show that $2^{\sqrt{2}}>1+\sqrt{2}$
From the given inequalities, I deduce that $(1+\sqrt{2})<\sqrt[5]{99}$ and $2^{\sqrt{2}}>2^{1.4}$. But I'm not sure on how to merge(if possible) the inequalities to get the desired result.

REPLY [8 votes]: Alternatively you can use Bernoulli's inequality $(1+x)^\alpha \geq 1+ \alpha x$, for $x >-1$ and $\alpha \geq 1$: $$2^\sqrt{2} = (1+1)^\sqrt{2} \geq 1 +\sqrt{2} \cdot 1 = 1 +\sqrt{2} $$