Inequality involving exponential of square roots How can I show that:
$$ 2e^\sqrt{3} \leq 3e^\sqrt{2} $$
? (that's all I have)
Thank you so much!
 A: Note that 
$$(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})=3-2=1$$
and
$$(\sqrt{3}+\sqrt{2})^2=5+2\sqrt{6}>5+2\cdot 2=9$$
Thus $\sqrt{3}+\sqrt{2}>3$, which gives $\sqrt{3}-\sqrt{2}<\frac{1}{3}$.
Now we claim that $e^{\frac{1}{3}}<\frac{3}{2}$, this is because
$$e<3<\frac{27}{8}=\left(\frac{3}{2}\right)^3$$
It follows that
$$e^{\sqrt{3}-\sqrt{2}}<e^{\frac{1}{3}}<\frac{3}{2}$$
hence
$$2e^{\sqrt{3}}<3e^{\sqrt{2}}$$
A: Divide both sides by $6$, which results in the more suggestive form
$$\frac{e^{\sqrt3}}{3}\le\frac{e^{\sqrt2}}{2}.$$
You now just have to check that the derivative of $e^{\sqrt x}/x$ is negative between $2$ and $3$.
A: Let $$f(x) =\frac{e^{\sqrt{x}}}{x}$$
Then $$f'(x)=\frac{e^\sqrt{x}\frac{1}{2\sqrt{x}}x-e^{\sqrt{x}}}{x^2}=\frac{e^{\sqrt{x}}}{x^2}(\frac{\sqrt{x}}{2}-1)$$
As $f'(x) <0$ on $(1,4)$ $f$ is decreasing and hence $f(2) >f(3)$.
A: Let $f(x)=\frac{e^{x}}{x^2}$. Then $$f'(x)=e^x\left(\frac{x-2}{x^3}\right)$$
So $f'(x)<0$ for $0<x<2$. Then $f(\sqrt{3})<f(\sqrt{2})$ and your result follows.
A: Take log both sides: $\ln 2 + \sqrt{3} \leq \ln 3 + \sqrt{2} \iff \dfrac{1}{2} \leq \dfrac{\ln \sqrt{3} - \ln \sqrt{2}}{\sqrt{3}-\sqrt{2}}$. 
Consider $f(x) = \ln x, x \in [\sqrt{2}, \sqrt{3}], f'(x) = \dfrac{1}{x}\geq \dfrac{1}{2}$.
The conclusion follows from MVT.
