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 $27$.
 A: Using Henry's approach, but with different estimates. If you know that the first three digits of $\ \sqrt{2}\ $ is $\ 1.41,\ $ then you have: $\ 1.40 < \sqrt{2} < 1.42857\ldots,\ $ i.e.,
$$\ \frac{7}{5} < \sqrt{2} < \frac{10}{7}.$$
So,
$$ \left( 9^{\sqrt{2}} \right)^7 < \left( 9^{ \frac{10}{7}} \right)^7 = 9^{10} = {81}^5 = \left( 80+1 \right)^5 $$
$$ = {80}^5 + 5\times {80}^4 + 10\times {80}^3 + 10\times {80}^2 + 5\times 80 + 1 $$
$$ < {80}^5 + 6 \times {80}^4 = 86 \times {80}^4 < 100 \times{80}^4 $$
$$ =100\times 8^4 \times {10}^4 < 4,100,000,000 = 4.1 \times 10^{9}, $$
whereas
$$ \left(\left(\sqrt{2}\right)^9\right)^7 = \left(\sqrt{2}\right)^{63} = 2^{31}\times \sqrt{2} > 1.4 \times 2^{31} = \frac{7}{2} \times 2^{31} $$
$$ = 7 \times 2^{30} = 7 \times \left( 2^{10} \right)^3 = 7 \times 1024^{3} > 7 \times 1000^3 = 7 \times 10^9. $$
A: It boils down to the comparison of some powers of $2$ and $3$, if I didn't do any mistake, in the following way: $9^{\sqrt{2}}<\sqrt{2}^9$ if $9<2^{\frac{9}{2\sqrt{2}}}$, by using the fact that $\sqrt{2}<1.415$, if $9<2^{\frac{9}{2\times 1.415}}$ if $9<2^{3.18}$ if $3^{100}<2^{159}$ which is true since:

*

*$2^{159}>\frac{1}{2}(1,02\times 10^{3})^{16}>\frac{1.32\times10^{48}}{2}=6,6\times10^{47}$.

*$3^{100}<(6\times 10^{4})^{10}=3^{10}\times 2^{10}\times 10^{40}<6\times10^{4}\times 1025\times 10^{40}=6,15\times 10^{47}$
A: $(\sqrt{2})^9 \quad vs.\quad 9^\sqrt{2}\qquad$// square both side
$2^9=512 \quad vs.\quad 
3^{4\sqrt{2}} \;≈\; 3^{5.657} \;<\; 3^5\,3^\frac{2}{3}$
$\displaystyle 3^5\,3^\frac{2}{3} = 243×2×\sqrt[3]{1+\frac{1}{8}} \;<\;486×\left(1+\frac{1}{3×8}\right) = 506.25$
$→ (\sqrt{2})^9 \;>\; 9^\sqrt{2}\qquad$
A: 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.
A: 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 $$9^{\sqrt{2}} \lt 9^{17/12}=3^{17/6} = 129140163^{1/6} < 134217728^{1/6} =2^{27/6} =2^{9/2} = \sqrt{2}^9$$
A: 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.
