# Show that $2^{\sqrt{2}}>1+\sqrt{2}$

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.

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

$$2^{1.4} = 2^{7/5} = (2^7)^{1/5} = 128^{1/5} > 99 ^{1/5}$$

• ah why couldn't I see that, thank you. Aug 1, 2015 at 0:50
• no problemo sir Aug 1, 2015 at 0:51

Work it out

$$\Big( 2^{\sqrt{2}} \Big)^5 > \Big( 2^{1.4} \Big)^5 = 2^7 > 99 > \Big( 1 + \sqrt{2} \Big)^5$$

What is $2^{1.4}$ to the $5$-th power? $1.4\times 5=7$ so $(2^{1.4})^5=2^7=128>99$.

Therefore, $2^{1.4}>\sqrt[5]{99}$ and you can finish from there.