How to prove $e^{\sqrt 2}>3$? 
How to prove $e^{\sqrt 2}>3$?

If I know the fact that $\sqrt{2}>\log(3)$, then the result follows (because $e^{x}$ is increasing). Can you give different solutions for this problem?
 A: Another approach uses the fact that for any positive $x$ and $n\rightarrow +\infty$, the expression
$(1+\frac{x}{n})^n$
strictly increases monotonically as a function of $n$ to $e^x$.  Then put in $x=\sqrt{2}, n=3$ to get
$e^{\sqrt{2}}>(1+\frac{\sqrt{2}}{3})^3$
$(1+\frac{\sqrt{2}}{3})^3=1+\sqrt{2}+\frac{2}{3}+\frac{2\sqrt{2}}{27}>1+\sqrt{2}+\frac{2}{3}>3$
where the last inequality in the chain is proved from $2>(16/9)$ thus $\sqrt{2}>(4/3)$.
A: For all $x$ such that $1 \leqslant x \leqslant 3$ we have $x(4 - x) = 4 - (x - 2)^2 \leqslant 3$, whence
$$
\log3 = \int_1^3\frac{dx}{x} \leqslant \int_1^3 \frac{4 - x}{3}\,dx = \frac{4}{3} < \sqrt{2}.
$$
A: $e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\cdots$, plug in $x=\sqrt{2}$.
A: $$\log(3) = \int_{1}^{3}\frac{dx}{x}\stackrel{CS}{\leq}\sqrt{\int_{1}^{3}\,dx\int_{1}^{3}\frac{dx}{x^2}}=\frac{2}{\sqrt{3}}<\sqrt{2}.$$
CS stands for Cauchy-Schwarz, of course.
A: If you know a few digits$^*$ of $e$ and $\sqrt3$, then you can quickly verify that $e^2 > 3\sqrt3$. Therefore
$$e^2>3^{3/2}>3^\sqrt2,$$
and raising both sides of this inequality to the power of $\frac{\sqrt2}2$ verifies that $e^\sqrt2>3$.

(*) In fact, we only need to know that $e > 2.5$ and $\sqrt3<2$, for then
$$e^2 > 2.5^2 > (2.5+5)(2.5-5) = 3\cdot2 > 3\sqrt3.$$
A: First, observe that for $x\geq0$, we have $e^x \geq 1+x+\frac{x^2}2$, so
$$\begin{align}
e^x - 2x &\geq \left(1+x+\frac{x^2}2\right) - 2x\\
&=\frac12\left(x^2-2x+2\right)\\
&=\frac12(x-1)^2 + 1\\
&>0.
\end{align}$$
Therefore,
$$\int_0^\sqrt2e^x\ \text dx \geq \int_0^\sqrt22x\ \text dx,$$
so
$$e^\sqrt2-1 \geq 2.$$
A: A well-known inequality states that $\log(1 + x) < x$ for $x > -1$ and $x \ne 0$. Taking $x = 1/4$,
$$
\log3 < \log\frac{3125}{1024} = \log\frac{5^5}{4^5} = 5\log\frac{5}{4} = 5\log\left(1 + \frac{1}{4}\right) < \frac{5}{4} < \sqrt{2}.
$$
