<s>[INST] 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!
 [/INST] 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
$$2e^{\sqrt{3}}<3e^{\sqrt{2}}$$
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