Proving an Inequality using a Different Method Is there another way to prove that: If $a,b\geq 0$ and $x,y>0$
$$\frac{a^2}{x} + \frac{b^2}{y} \ge \frac{(a+b)^2}{x+y}$$
using a different method than clearing denominators and reducing to $(ay-bx)^2 \ge 0$? 
 A: That is Titu's lemma, that is equivalent to Cauchy-Schwarz inequality:
$$ (x+y)\left(\frac{a^2}{x}+\frac{b^2}{y}\right)\geq(a+b)^2. $$
A: $\frac{a^2}{x} + \frac{b^2}{y} 
\ge \frac{(a+b)^2}{x+y}
$
Let
$u = a^2/x$
and
$v = b^2/y$,
so
$a = \sqrt{ux}$
and
$b = \sqrt{vy}$.
The inequality then becomes
$u+v
\ge \frac{(\sqrt{ux}+\sqrt{vy})^2}{x+y}
= \frac{ux+vy+2\sqrt{uxvy}}{x+y}
$
or
$ux+uy+vx+vy
\ge ux+vy+2\sqrt{uxvy}
$
or
$uy+vx
\ge 2\sqrt{uxvy}
$
or
$(\sqrt{uy}-\sqrt{vx})^2
\ge 0
$.
I don't know different this really is,
since it does clear the denominator,
but it looks different.
Anyway,
most inequalities usually 
are equivalent to
$(something)^2 \ge 0
$. 
A: One method of solving any inequality is to determine points where the associated equality is true or where the functions involved are not continuous.  That is because, as long as $a$ and $b$ are continuous, we can only go from "$a> b$" to "$a< b$", or vice-versa, by going through "$a= b$".
However here, the simplest way to solve the equality is by "clearing denominators" so apparently you would not consider that a "different" way.
