Show $ \left| \frac {xy}{x^2-xy+y^2} \right|\le 1 , \forall (x,y) \neq (0,0).$

Question

My approach: $$ \left| \frac {xy}{x^2-xy+y^2} \right|= \frac {\lvert xy \rvert}{\lvert (x-y)^2+xy\rvert}\leq \frac {\lvert {xy} \rvert}{\lvert {xy} \rvert} = 1$$

Is this right?

If yes, how do we know that $\lvert (x-y)^2 +xy\rvert \geq \lvert xy\rvert$ ?

Answer 1

We have \begin{align} x^2+y^2\geq 0 &\text{ and } (x-y)^2\geq 0\\ \implies x^2+y^2-xy\geq -xy &\text{ and } (x-y)^2+xy\geq xy\\ \implies x^2+y^2-xy\geq -xy &\text{ and } x^2+y^2-xy\geq xy\\ \implies x^2+y^2-xy\geq |xy|\\ \implies |x^2+y^2-xy|\geq |xy| \end{align}

Answer 2

You can also use $2|xy|\leq x^2+y^2$ which follows from AM-GM (or you can verify it directly by considering $|xy|=\pm xy$). Note that the denominator is positive for $(x,y)\ne (0,0)$. So $$\left|\frac{xy}{x^2-xy+y^2}\right|\leq\frac{1}{2}\frac{x^2+y^2}{x^2-xy+y^2} $$ and now you have to show $$\frac{1}{2}\frac{x^2+y^2}{x^2-xy+y^2}\leq 1 \Leftrightarrow \\ \frac{x^2+y^2}{x^2-xy+y^2}\leq 2 \Leftrightarrow \\ 0\leq x^2-2xy+y^2 $$ which is true. Equality is reached only if $x=y$.

Answer 3

I think the following reasoning a bit of better.

If $xy=0$ so it's obvious.

But for $xy\neq0$ we obtain: $$\left|\frac{xy}{x^2-xy+y^2}\right|=\frac{|xy|}{x^2-xy+y^2}\leq\frac{|xy|}{x^2-|xy|+y^2}=\frac{|xy|}{(|x|-|y|)^2+|xy|}\leq\frac{|xy|}{|xy|}=1.$$