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This question is an exact duplicate of:

How do I prove the following inequality :

$$\Bigg(\frac{2}{\alpha^2} \, \big( e^{\alpha x} - e^{\alpha y} \big) \, + \, e^{\alpha y} (y^2 - x^2) \; \Bigg) \geq 0 $$

given, $x, y \geq 0$ ?

Can anyone provide me with hints about this problem ?

Here, $\alpha$ is a strictly positive constant.

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marked as duplicate by Did, Daniel W. Farlow, choco_addicted, R_D, M. Vinay Jun 16 '16 at 6:01

This question was marked as an exact duplicate of an existing question.

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    $\begingroup$ But it's false when $x=y$. $\endgroup$ – John Wayland Bales Jun 16 '16 at 3:55
  • $\begingroup$ It's a continuous surface. If you add the restriction that $x\ne y$ or change the inequality to $\ge0$ perhaps you can find the critical points of the surface where the first partials are both $=0$ in the first quadrant and apply the second derivative test. If the critical points are minima and have a non-negative value then you are done. A lot of ifs, but it is an approach. $\endgroup$ – John Wayland Bales Jun 16 '16 at 4:20
  • $\begingroup$ This is obviously false, try $x=y+1$ when $y\to\infty$. $\endgroup$ – Did Jun 16 '16 at 5:20
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Well,the simplest thing to do-take the natural logarithm of both sides-doesn't work here since In (0) is undefined.

However, it appears the equation is separable-i.e. the variable terms can be rearranged so that x terms are on one side and the y terms are on the other. So let's do that.

$$\Bigg(\frac{2}{\alpha^2} \, \big( e^{\alpha x} - e^{\alpha y} \big) \, + \, e^{\alpha y} (y^2 - x^2) \; \Bigg) \geq 0 $$ $\rightarrow$ $\frac{2}{\alpha^2}e^{\alpha x} - \frac{2}{\alpha^2}e^{\alpha y} + e^{\alpha y} y^2 - e^{\alpha y}x^2 \geq 0$ $\rightarrow$

$\frac{2}{\alpha^2}e^{\alpha x} - e^{\alpha y}x^2 \geq \frac{2}{\alpha^2}e^{\alpha y} + e^{\alpha y} y^2 $ = $\frac{2}{\alpha^2}e^{\alpha x} - e^{\alpha y}x^2 \geq e^{\alpha y} (\frac{2}{\alpha^2}+ y^2 )$

Multiply both sides through by $e^{-\alpha y}$.

$\frac{2}{\alpha^2}e^{\alpha (x -y)} - x^2 \geq \frac{2}{\alpha^2}+ y^2 $

Rearranging again:

$\frac{2}{\alpha^2}(e^{\alpha (x -y)} - 1) \geq x^2 + y^2 $

Clearly $x\neq y$ because then the entire left hand side of the original equation equals 0. Substituting polar coordinates gives:

$\frac{2}{\alpha^2}(e^{\alpha r(cos{\theta} -sin{\theta})} - 1) \geq r(cos ^2{\theta} + sin^2 {\theta}) =r^2$

$r^2$ is the square of the radius of a circle in the plane, which by definition is greater then or equal to 0. That should do it.

Hope it helps. Good luck!

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  • $\begingroup$ $\frac{2}{\alpha^2}e^{\alpha x} - e^{\alpha y}x^2 \geq \frac{2}{\alpha^2}e^{\alpha y} + e^{\alpha y} y^2 $ = $\frac{2}{\alpha^2}e^{\alpha x} - e^{\alpha y}x^2 \geq e^{\alpha y} (\frac{2}{\alpha^2}+ y^2 )$. I don't think that there will be a "+" sign before the expression $e^{\alpha y} y^2$ in this line. $\endgroup$ – Dwaipayan Gupta Jun 16 '16 at 8:28

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