<s>[INST] Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
 [/INST] The given inequality can be transformed to various equivalent forms:
$$\begin{align}
9^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^9 
&\iff (3^2)^{\sqrt{2}} \stackrel{?}{<} \sqrt{2}^{3\times 3} 
\iff 3^{2\sqrt{2}} \stackrel{?}{<} {(2\sqrt{2})}^3\\
&\iff 3^{\frac13} \stackrel{?}{<} {(2\sqrt{2})}^{\frac{1}{2\sqrt{2}}}\end{align}$$
Notice $3 > 2\sqrt{2} \sim 2.828 > e \sim 2.718 $ and the function $x^{\frac1x}$ is strictly decreasing for $x > e$ (standard calculus exercise). The rightmost "inequality" is true
and hence the original "inequality" $9^{\sqrt{2}} < \sqrt{2}^9$ is true.
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