JOMO 7, Short 5

We have, $\frac{x^2+y^2+2xy}{x^2+y^2}$ $=1+\frac{2xy}{x^2+y^2}$ Now we know that square of a number is always greater than or equal to 0, $\implies (x-y)^2\geq 0 \implies x^2+y^2\geq 2xy$ Back to our expression, there will be the maximum value if $x^2+y^2=2xy$ i.e. the maximum value is $1+1=\boxed2$

By Cauchy Schwarz,

$(x^2 + y^2)(1+1) \geq (x+y)^2$

$2 \geq \dfrac{(x+y)^2}{x^2 + y^2}$

Therefore maximum value is $\boxed{2}$ when $\frac{x}{1} = \frac{y}{1}$

Another technique is to use the polar coordinate transformations: x = r cos(k), y = r sin(k), x^2 +y^2 = r^2. Substituting these into the original expression gives:

[r cos(k) + r sin(k)]^2 / r^2 = r^2 * [cos(k) + sin(k)]^2 / r^2 = [cos(k) + sin(k)]^2 = [cos(k)]^2 + [sin(k)]^2 + 2 cos(k) sin(k);

or sin(2k) + 1 (i).

Hence, the maximum value of (i) is just 2.

I did the same way as sanjit sir did:-)