Inequality Problem involving exponential expression How do I prove the following inequality :
$$\frac{2}{\alpha^2} \bigg( e^{\alpha y} - e^{\alpha x} \bigg) +  e^{\alpha x} \bigg( x^2 - y^2 \bigg) > 0 \; \;?$$
Here, $\alpha > 0, y < x$.
Additional information : both $x$ and $y$ are strictly greater than 0 !
 A: This is false.
Take $x = 0$ and $y = -\frac{1}{\alpha}$.  Then
$$\frac{2}{\alpha^2} \left( e^{\alpha y} - e^{\alpha x} \right) +  e^{\alpha x} \left( x^2 - y^2 \right) =  \frac{2}{\alpha^2} \left( e^{-1} - 1\right) + 1 \left(0 - \frac{1}{\alpha^2}\right)$$
Which simplifies to
$$ \frac{1}{\alpha^2} \left( 2 e^{-1} - 3 \right) $$
Which is negative
>>> 2 * exp(-1) - 3
-2.2642411176571153

Perhaps you are missing another constraint?
A: It can be rewritten as:

Of which - under the constraints you specify - the leftmost two terms together are always < 0, and the rightmost two terms together are always > 0. 
Which means that as long as the two rightmost terms together are larger than the two leftmost terms, you're fine. 
Despite the fact that specific cases can be found for which this condition isn't met (as answered before, and proving the inequality as it is is false) there clearly are cases for which this constraint is met. For example $a = \sqrt2, x = 2, y = 1$. This will get you closer to finding that missing constraint which @Matthew Drury mentioned. 
