MHB Can Jensen's Inequality Solve the Inequality Challenge III?

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The discussion focuses on proving the inequality \( e^{1/e} + e^{1/\pi} \geq 2e^{1/3} \) using the AM-GM inequality. It demonstrates that \( e^{1/e} + e^{1/\pi} \) can be bounded below by \( 2\sqrt{e^{1/e}e^{1/\pi}} \), leading to the conclusion that this expression exceeds \( 2e^{1/3} \). Additionally, a reference is made to using Jensen's inequality as an alternative method to solve the problem. The participants engage in mathematical reasoning to validate the inequality, emphasizing the effectiveness of both AM-GM and Jensen's approaches. The discussion highlights the interplay between different mathematical techniques in addressing inequality challenges.
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Show that $e^\dfrac{1}{e}_{\phantom{i}}+e^{\dfrac{1}{\pi}}_{\phantom{i}} \ge2e^{\dfrac{1}{3}}_{\phantom{i}}$.
 
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anemone said:
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]
 
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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:

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.
 
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