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I'm having a bit of trouble showing the following problem :

Show that : $x^4 - 2x^2y^2 + y^2 \geq 0$ for $x\in D$ where D is a domain around $O(0,0)$.

I guess that for $|x| \leq1/2$ it holds : $$x^4-2x^2y^2 + y^2 \geq \frac{1}{2}y^2$$

which means that $x^4-2x^2y^2 + y^2 \geq 0$ for $|x| < 1/2$, but I can't see how I could prove that.

Excuse me if it's elementary, sometimes I'm a bit blind on inequality proofs.

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Since \begin{align*} x^4-2x^2y^2+y^2&=x^4-2x^2y^2+y^4+y^2-y^4\\ &=\left(x^2-y^2\right) ^2+y^2-y^4\end{align*}

it follows that $x^4-2x^2y^2+y^2\ge0\;\Longleftarrow\;y^2-y^4\ge 0\;\iff\;|y|\le1$.

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  • $\begingroup$ The first equivalence $\iff$ in the last line is actually only a reverse implication $\impliedby$. $\endgroup$ – Did Nov 22 '17 at 9:33
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We are in a domain around $(0,0)$, in particular $y <1$ and $(1-y^2)>0$, so \begin{eqnarray*} x^4-2x^2y^2+y^2 = (x^2-y^2)^2 +y^2(1-y^2) > 0 . \end{eqnarray*}

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It's $$(x^2-y^2)^2+y^2-y^4\geq0,$$ which is true for $-1\leq y\leq1$.

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