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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.

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4 Answers 4

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

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

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  • $\begingroup$ ah why couldn't I see that, thank you. $\endgroup$ Aug 1, 2015 at 0:50
  • $\begingroup$ no problemo sir $\endgroup$
    – muaddib
    Aug 1, 2015 at 0:51
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Work it out

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

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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.

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