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The inequality is the following : $\frac{x^2+y^2}{4} \leq e^{x+y-2}$, where $x,y \geq0$.

I tried manipulating the inequality using Taylor series, but I couldn't find a conclusive result . Any ideas ?

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Hint: $x=\rho \cos \theta$, $y=\rho \sin \theta$, with $\theta \in [0;\dfrac{\pi}{2}]$ and $\rho \geq 0$

Then $\dfrac{x^2+y^2}{4}=\dfrac{\rho^2}{4}$ and $e^{x+y-2}=\dfrac{e^{\rho (\cos \theta + \sin \theta)}}{e^2}$

Can you conclude something by comparing the two expression written as such?

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  • $\begingroup$ Not really, can you explain ? $\endgroup$
    – vernon
    Commented Feb 9, 2015 at 14:53
  • $\begingroup$ does it concern the minima of cos(theta)+sin(theta) on 0;pi/2 > $\endgroup$
    – vernon
    Commented Feb 9, 2015 at 14:57
  • $\begingroup$ Got it, thans you $\endgroup$
    – vernon
    Commented Feb 9, 2015 at 15:04

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