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!

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.
