Proving an Inequality using a Different Method

Question

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 $(ay-bx)^2 \ge 0$?

Answer 1

That is Titu's lemma, that is equivalent to Cauchy-Schwarz inequality: $$ (x+y)\left(\frac{a^2}{x}+\frac{b^2}{y}\right)\geq(a+b)^2. $$

Answer 2

$\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 $.

Answer 3

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.