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 $27$.

What you want is a suitable rational upper bound  to $\sqrt{2} \approx 1.4142$ such as $\frac{17}{12} \approx 1.4167$ which you can find for example with continued fractions. As a check, $\left(\frac{17}{12}\right)^2=\frac{289}{144}>2$.
Then you can say $$9^{\sqrt{2}} \lt 9^{17/12}=3^{17/6} = 129140163^{1/6} < 134217728^{1/6} =2^{27/6} =2^{9/2} = \sqrt{2}^9$$
