Q12 (b)

Criteria	Marks
Provides correct proof	2
Expands the numerator of the LHS and splits into two fractions, or equivalent merit	1

Sample answer:

$\frac{(x+y)^2}{x^2+y^2}=\frac{x^2+2xy+y^2}{x^2+y^2}=1+\frac{2xy}{x^2+y^2}$

But
$\frac{a+b}{2}\ge\sqrt{ab},\quad\text{therefore}\quad\frac{x^2+y^2}{2}\ge\sqrt{x^2y^2}$
and
$x^2+y^2\ge2|xy|\ge2xy$
(since $\sqrt{x^2}=|x|$ and here we have $a,b\ge0$).

$1\ge\frac{2xy}{x^2+y^2}$
$\therefore\frac{(x+y)^2}{x^2+y^2}\le1+1\le2$

Alternative solution

$(x-y)^2\ge0\qquad\forall x,y\in\mathbb R$
$x^2-2xy+y^2\ge0$
$2x^2-2xy+2y^2\ge x^2+y^2$
$2(x^2+y^2)\ge x^2+y^2+2xy\ge(x+y)^2$
$\therefore2\ge\frac{(x+y)^2}{x^2+y^2}$