Prove $\frac{x^2}{a} + \frac{y^2}{b} \geq \frac{(x + y)^2}{a + b}$, with $x, y, a, b \in \mathbb{R}$ and $a, b > 0$

Question

Prove $\frac{x^2}{a} + \frac{y^2}{b} \geq \frac{(x + y)^2}{a + b}$, with $x, y, a, b \in \mathbb{R}$ and $a, b > 0$

Proof:

$\frac{x^2}{a} + \frac{y^2}{b} \geq \frac{(x + y)^2}{a + b}$

$\iff (bx^2 + ay^2)(a + b) \geq ab(x^2 + 2xy + y^2)$

$\iff abx^2 + a^2y^2 + b^2x^2 + aby^2 \geq abx^2 + 2abxy + aby^2$

$\iff a^2y^2 + b^2x^2 \geq 2abxy$

$\iff (ay - bx)^2 \geq 0$

This result prompted me to use the Cauchy-Schwarz inequality to prove the statement directly $LHS \geq RHS$. However, I couldn't seem to figure out a solution in this manner. I wondered if somebody here has a solution that doesn't involve going backward like above.

Answer 1

Direct Cauchy-Schwarz would be a single step

$$({x\over\sqrt{a}}\sqrt{a}+{y\over\sqrt{b}}\sqrt{b})^2\le({x^2\over a}+{y^2\over b})(a+b)$$

Simplification gives the required inequality

Answer 2

Let Cauchy–Bunyakovsky–Schwarz run over $$(x + y)^2\:=\: \left(\sqrt{a}\,\frac{x}{\sqrt a} \,+\, \sqrt{b}\,\frac{y}{\sqrt b}\right)^2$$ (and eventually divide by $a+b\,$).