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I am working on the same question as a poster here and I was able to prove the inequality by re-factoring to:

$$ (x-z)^2 + 4y(2-x-z) + 4y^2 \geq 0$$

and arguing that given the conditions, this holds (first and third terms are positive, for the second $y>0$ by definition and so is $2-x-z$).

However, I am also asked to determine when the equality holds. Given that I have a sum, I don't see a way to do that:

$$ (x-z)^2 + 4y(2-x-z) + 4y^2 = 0$$

Can someone explain the general direction I should move in to be able to prove the equality? The only bright thought was dividing by $4y$ and looking for a substitution, but this did not yield anything.

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  • $\begingroup$ @J.M., thanks for the edit, I was just searching through help how to make the formulas appear properly! $\endgroup$
    – jojxvwyv
    Jan 8, 2012 at 4:29

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You've just argued that all term are non-negative real numbers, yes? Ergo, their sum must be strictly positive if any are strictly positive. So $y=0$ and $x=z$ is a necessary and sufficent condition.

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Well, if you get each term to equal zero, then the entire thing will equal zero. If we have that $y=0$ and $x-z=0$, then equality will hold. Then equality holds when $y=x-z=0$. There should be other conditions for which equality holds, but this is one.

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