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How can I show that:

$$ 2e^\sqrt{3} \leq 3e^\sqrt{2} $$

? (that's all I have)

Thank you so much!

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  • $\begingroup$ I'm sure you can do this without calculus, just starting with the fact that $e^x\ge x+1$ (with equality iff $x=1$). EDIT: Just use Frank Lu's answer; to prove that $e^{1/3}<\frac32$, just set $x=-\frac13$ in my equation. $\endgroup$ – Akiva Weinberger Feb 27 '15 at 5:26
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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.

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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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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 $3$.

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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 $f(2) >f(3)$.

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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.

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