Proving Absolute Value Inequality Using Lagrange Multipliers 
a) Use Lagrange multipliers method to find the maximum and minimum
  values of the function $$f(x,y) = xy$$ over the curve $$x^2-yx+y^2 =1$$
  b)Use the result of the preceding part, or otherwise, to prove that
  $$\left| \frac{xy}{x^2-yx+y^2 }\right| \leq 1, \forall(x,y) \neq 0$$

For part a) I got the maximum value to be $1$ and the minimum value to be $-1/3$, however I have no idea how to use that to do part b) since those values only apply when $x^2-yx+y^2 =1$. I have tried computing $\nabla f$ hoping that the maximum and minimum values of $f$ lie on $x^2-yx+y^2 =1$ but that is not the case. I don't have any other ideas on how to approach this.
Thanks for any help. 
 A: Note that $$\left| \frac{xy}{x^2-yx+y^2 }\right| \leq 1, \forall(x,y) \neq 0 \iff -1\le\frac{xy}{x^2-yx+y^2 }\le 1$$
Multiply by the positive expression $x^2-yx+y^2$ to get $$-x^2+yx-y^2 \le xy \le x^2-yx+y^2$$
Both inequalities are simple to verify.  
A: Suppose we have $x$ and $y$, such that $x^2-yx+y^2\neq 1$. You can prove that $x^2-yx+y^2\geq 0$, and that $x^2-yx+y^2=0$ if and only if $(x,y)=(0,0)$ (which is a case we don't care about anyway, in the problem statement).  So, we know that $x^2-yx+y^2>0$.
Say $x^2-yx+y^2=c^2$, and define $\bar{x}=x/c$ and $\bar{y}=y/c$.  Then we have
$$
\bar{x}^2-\bar{y}\bar{x}+\bar{y}^2=\frac{x^2-yx+y^2}{c^2}=1,
$$
and further
$$
\frac{xy}{x^2-yx+y^2}=\frac{\bar{x}\bar{y}}{\bar{x}^2-\bar{y}\bar{x}+\bar{y}^2}=\bar{x}\bar{y}.
$$
Can you see how to finish from here?
A: Since $$\left|\frac{x}{y}+\frac{y}{x}\right|\geq2,$$ by the triangle inequality we obtain:
$$\left|\frac{xy}{x^2-xy+y^2}\right|=\frac{1}{\left|\frac{x}{y}+\frac{y}{x}-1\right|}\leq\frac{1}{\left|\frac{x}{y}+\frac{y}{x}\right|-1}\leq\frac{1}{2-1}=1.$$
A: The max is indeed $f(-1,-1)=f(1,1)=1$, but min is $f\left(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)=f\left(\frac1{\sqrt{3}},-\frac1{\sqrt{3}}\right)=-\frac13.$
Hence: $-\frac13\le xy\le 1 \Rightarrow |xy|\le 1.$
Note that $x^2-xy+y^2=1$ is the level curve. If $x^2-xy+y^2=k>0$, then $-\frac k3\le xy\le k \Rightarrow |xy|\le k$. 
Hence:
$$\left| \frac{xy}{x^2-yx+y^2 }\right|= \left| \frac{xy}{k}\right|\le \left|\frac kk\right|=1.$$
