<s>[INST] If $\frac{x^2+y^2}{x+y}=4$, then what are the possible values of $x-y$? 
If $\frac{x^2+y^2}{x+y}=4$,then all possible values of $(x-y)$ are given by
$(A)\left[-2\sqrt2,2\sqrt2\right]\hspace{1cm}(B)\left\{-4,4\right\}\hspace{1cm}(C)\left[-4,4\right]\hspace{1cm}(D)\left[-2,2\right]$

I tried this question.
$\frac{x^2+y^2}{x+y}=4\Rightarrow x+y-\frac{2xy}{x+y}=4\Rightarrow x+y=\frac{2xy}{x+y}+4$
$x-y=\sqrt{(\frac{2xy}{x+y}+4)^2-4xy}$, but I am not able to proceed. I am stuck here. Is my method wrong?
 [/INST] Given $$\displaystyle \frac{x^2+y^2}{x+y} = 4\Rightarrow x^2+y^2 = 4x+4y$$
So we get $$x^2-4x+4+y^2-4y+4 = 8\Rightarrow (x-2)^2+(y-2)^2 = 8$$
Now we can write $$x-y = (x-2)-(y-2) = \left[(x-2)+(2-y)\right]$$
Now Using $\bf{Cauchy\; Schwartz\; Inequality}$
$$\displaystyle \left[(x-2)^2+(2-y)^2\right]\cdot [1^2+1^2]\geq \left[x-2+2-y\right]^2$$
So $$8\times 2 \geq (x-y)^2$$
So $$(x-y)^2\leq 4^2$$
So $$-4 \leq (x-y)\leq 4\Rightarrow x-y\in \left[-4,4\right]$$
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