Can Jensen's Inequality Solve the Inequality Challenge III?

[anemone]

Show that $e^\dfrac{1}{e}_{\phantom{i}}+e^{\dfrac{1}{\pi}}_{\phantom{i}} \ge2e^{\dfrac{1}{3}}_{\phantom{i}}$.

 

[Opalg]

Show that $e^{1/e}+e^{1/\pi} \geqslant 2e^{1/3}$.

[sp]AM-GM: $e^{1/e}+e^{1/\pi} \geqslant 2\sqrt{\mathstrut e^{1/e}e^{1/\pi}} = 2\exp\Bigl(\frac12\bigl(\frac1e + \frac1\pi\bigr)\Bigr)$. By AM-GM again, $\frac12\bigl(\frac1e + \frac1\pi\bigr) \geqslant \sqrt{\frac1{e\pi}}.$ but $e<11/4$ and $\pi < 16/5$, so $e\pi <176/20 <9.$ Thus $\sqrt{e\pi} <3$, and $\sqrt{\frac1{e\pi}} > \frac13$. Therefore $2\exp\Bigl(\frac12\bigl(\frac1e + \frac1\pi\bigr)\Bigr) > 2e^{1/3}.$[/sp]

 

[anemone]

Thank you Opalg for participating!

Here is another method(not my own product) that used the Jensen inequality as the main weapon to crack this problem:

Spoiler

Consider the function $f(x)=e^{\dfrac{1}{x}}_{\phantom{i}}$ for $x>0$.

We have $f'(x)=-\dfrac{\dfrac{1}{x}}{e^{\dfrac{1}{x}}_{\phantom{i}}}<0$, and $f''(x)=e^{\dfrac{1}{x}}_{\phantom{i}}\left(\dfrac{2}{x^3}+\dfrac{1}{x^4} \right)>0$, hence $f$ is decreasing and convex.

By the Jensen Inequality formula, we have

$\dfrac{1}{2}(f(e)+f(\pi))\ge f\left( \dfrac{e+\pi}{2} \right)$

On the other hand, we have $\dfrac{e+\pi}{2}<3$ and since $f$ is decreasing, $f\left( \dfrac{e+\pi}{2} \right)>f(s)$ and from here the result follows.