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I have the multivariable function :

$$f_{\alpha}(x,y)=\frac{\left | xy \right |^{\alpha}}{x^{2}-xy+y^2}$$

I think that an upper bound in $(0,0)$ for $\alpha > 1$ is : $$\left | f_{\alpha}(x,y) \right | \leq \frac{\left | xy \right |^{\alpha}}{\left | xy \right |}=\left | xy \right |^{\alpha-1}$$

But to do so I have to prove that :

$\left | (a-b)^{2}+ab \right | \geq \left | ab \right |$

I've used a graphical study to show it but an inequality analysis would be much better :-/ Do you have any idea of the way to prove this inequality ?

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    $\begingroup$ need to put dollar signs around your LateX commands. $\endgroup$ – Alan Oct 12 '14 at 4:56
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As $(a-b)^2+ab=a^2-ab+b^2= (a-\frac{b}{2})^2+\frac{3}{4}b^2 \geq 0$ the inequality is

$$a^2-ab+b^2 \geq |ab|$$

This is easy to prove when $ab \geq 0$ and trivial when $ab <0$.

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$|a^2-ab+b^2|\geq |ab|$

  • $a\ \text{and}\ b\ \text{have different sign}\rightarrow a^2+b^2+ab\geq ab$
  • $a\ \text{and}\ b\ \text{have same sign}\rightarrow a^2+b^2-ab\geq ab\rightarrow a^2-2ab+b^2\geq0\rightarrow (a-b)^2\geq 0$
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Dividing both sides by $|ab|$, your statement is equivalent to $$\left| \frac{(a-b)^2+ab}{ab}\right|=\left| \frac ab+\frac ba-1 \right|\geq1$$ By noting that $\left|x+\frac1x\right|\geq2$ (I leave it to you to show this), the above inequality is seen to be true, and therefore so was your original statement.

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