$2^\sqrt{10}$ vs $3^2$

Is there a neat way to show that $2^\sqrt{10} < 3^2$?

I have tried raising to larger powers, like $(2^\sqrt{10})^{100}$ vs $3^{200}$ but the problem is the two functions $2^{x\sqrt{10}}$ and $3^{2x}$ are almost equivalent, and there is no point (that I can find) where one function is "obviously" larger than the other.

Looking at $2^{2\sqrt{10}}$ vs $3^4$ I tried to find a way of showing that $2^{2\sqrt{10}} < 2^6+2^4 = 2^4(2^2+1)$ but couldn't see any neat solution.

Any help or hints are appreciated.

edit: I should have specified when I say "neat solution" I was looking for a method readily done by hand. I realise this might be an unrealistic limitation, but it was why I'm interested.

Final thoughts before going to bed: I was looking at the functions $f(x)=2^{\sqrt{1+x^2}}$ and $g(x)=x^2$. At $x=0$, clearly $f>g$. They are equal at $x=2\sqrt{2}$. At $x=4$, again $f>g$. This mean that somewhere in the interval $2<x<4$, $g>f$ (as they are both convex). The task then is to try to find a point such that $x>3$ and $g>f$. But then another fun inequality pops out... $2^{\sqrt{11}}$ vs $10$...

Solution 1:

$$2^{\sqrt{10}}=8^{\sqrt{1+\frac{1}{9}}}<8^{1+\frac{1}{18}}=2^{\frac{19}{6}}<9.$$ Because $$2^{19}=2(500+12)^2=524288<531441=(700+29)^2=3^{12}.$$ There is another way to show a last inequality, but it's not so nice: $$2^{19}<3^{12}$$ it's $$\left(1+\frac{1}{8}\right)^6>2,$$ which is true because $$\left(1+\frac{1}{8}\right)^6>1+\frac{6}{8}+\frac{15}{64}+\frac{20}{512}>2.$$

Solution 2:

It follows from the maclaurian series of the function $\sqrt{k^2+h}$ for $|h|\leq k$ that , $$\sqrt{k^2+h}<k+\frac{h}{2k}$$ Put $k=3$ and $h=1$ we get , $$\sqrt{10}<3+\frac{1}{2\cdot 3}=3+\frac{1}{6}$$ Hence we must have $$2^{\sqrt{10}}<2^3\cdot \sqrt[6]{2}=8\cdot \sqrt[6]{2}$$ Therefore we only need to show that $$8\cdot \sqrt[6]{2}<3^2$$ Or in other words we need to show that $$\sqrt[6]{2}<\frac{9}{8}=1+\frac{1}{8}$$ or in other words if we raise both sides to $6th$ power ,we need to show that $$2<\left( \frac{9}{8}\right)^6=\left(1+\frac{1}{8}\right)^6$$ Expanding right hand side using binomial theorem we're left to prove that $$2<\sum \limits_{k=0}^6 {}^6C_k \cdot \frac{1}{8^k}$$ As the first term in the expansion on right hand side is obviously 1 therefore subtracting 1 both sides leaves us to prove that $$1<\sum \limits_{k=1}^6 {}^6C_k \cdot \frac{1}{8^k}$$ Now it's working time we don't need to know the full expansion. We'll just leave where the rhs exceeds 1.
Now ,$${}^6C_1\cdot \frac{1}{8}=\frac{6}{8}=\frac{3}{4}$$ $${}^6C_2\cdot \frac{1}{8^2}=\frac{30}{128}=\frac{15}{64}$$ $${}^6C_3 \cdot \frac{1}{8^3}=\frac{20}{512}=\frac{5}{128}$$ let's check now whether this exceeds 1 or not. Adding we get $$\frac{3}{4}+\frac{15}{64}+\frac{5}{128}=\frac{63}{64}+\frac{5}{128}=\frac{131}{128}=\frac{128+3}{128}=1+\frac{3}{128}>1$$ (Yay!! ;) ) Therefore the inequality is proven. Hence we have $$2^{\sqrt{10}}<3^2$$

Solution 3:

\begin{gather*} 2\times 19^2 = 2 \times 361 = 722 > 720 = 5 \times 12^2, \ \therefore\ \frac{19}{12} > \sqrt{\frac52}; \\ 2^{10} = 1024 < 1025 = 25 \times 41; \\ 3^6 = 729 > 725 = 25 \times 29; \\ 2 \times 29^2 = 2 \times 841 = 1682 > 1681 = 41^2; \\ \therefore\ 2 \times 3^{12} > 25^2 \times (2 \times 29^2) > 25^2 \times 41^2 > 2^{20}, \ \therefore 3^{12} > 2^{19}; \\ \therefore\ \log_23 > \frac{19}{12} > \sqrt{\frac52}, \ \therefore \log_23^2 > \sqrt{10}, \ \therefore 3^2 > 2^{\sqrt{10}}. \end{gather*}

"Another fun inequality" was recently added to the question.

Just for fun, then (because the new inequality cannot be considered as part of the question, this cannot be considered as part of an answer, but it's too long for a comment): \begin{gather*} 5^5 = 3125 > 3072 = 12 \times 256 = 12 \times 2^8, \ \therefore\ 5^{13} > 12 \times 10^8; \\ 29^3 = (30 - 1)^3 = 30^3 - 3 \times 30^2 + 3 \times 30 - 1 = 27000 - 2700 + 90 - 1 \\ = 24389 < 24400 = 61 \times 400, \ \therefore\ 29^6 < 61^2 \times 2^4 \times 10^4; \\ 61^2 = 3721 < 3750 = 6 \times 5^4, \ \therefore\ 29^6 < 6 \times 10^8, \ \therefore\ 5^{13} > 2 \times 29^6; \\ 2^{10} = 1024 < 1025 = 25 \times 41, \ \therefore\ 2^{60} < 5^{12} \times 41^6; \\ 41^2 = 1681 < 1682 = 2 \times 841 = 2 \times 29^2, \ \therefore\ 2^{60} < 5^{12} \times 2^3 \times 29^6; \\ \therefore\ 2^{60} < 5^{12} \times 2^2 \times 5^{13}, \ \therefore 2^{58} < 5^{25}, \ \therefore\ \log_25 > \frac{58}{25}; \\ 83^2 = 6889 > 6875 = 11 \times 625 = 11 \times 25^2, \ \therefore\ \sqrt{11} < \frac{83}{25}; \\ \therefore\ \sqrt{11} < 1 + \log_25 = \log_2{10}, \ \therefore\ 2^{\sqrt{11}} < 10. \end{gather*}

Solution 4:

A trick is to blow up the gap

$[2^\sqrt{10},\;3^2] → [2^{\sqrt{10}-3},\;9/8] → [2,\;(9/8)^{\sqrt{10}+3}]$

$1.26^3 = 2.000376$

$(9/8)^2 = 81/64 = 1+1/4+1/16 > 1.26$
$(9/8)^{\sqrt{10}+3} > (9/8)^6 > 1.26^3 > 2\qquad\qquad$ ⇒ RHS is bigger