<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] Remark: @achille hui posted a similar proof. But we got them  independently.
The desired inequality is written as
$$3^{2\sqrt 2} < (2\sqrt 2)^3$$
or
$$2\sqrt 2\, \ln 3 < 3\ln (2\sqrt 2)$$
or
$$\frac{\ln 3}{3} < \frac{\ln(2\sqrt 2)}{2\sqrt 2}. \tag{1}$$
Let
$$f(x) := \frac{\ln x}{x}.$$
We have
$$f'(x) = \frac{1 - \ln x}{x^2}.$$
Thus, $f'(x) < 0$ on $(\mathrm{e}, \infty)$.
Since $\mathrm{e} < 2\sqrt 2 < 3$, (1) is true.
We are done.
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