# Inequality: $2\sqrt{xz}+2\sqrt{yz}+2\sqrt{xy}\geq 3x+3y+3z-3$

Let $x,y,z$ be nonnegative real numbers satisfying $x^2+y^2+z^2=1$. Prove that:

$2\sqrt{xz}+2\sqrt{yz}+2\sqrt{xy}\geq 3x+3y+3z-3$

I have tried squaring both sides to get:

$8x\sqrt{yz}+8y\sqrt{xz}+8z\sqrt{xy}\geq14xy+14xz+14yz-18x-18y-18z+18$

(managed to use the condition $x^2+y^2+z^2=1$, but seems to get even more complicated)

Thanks for any help!

## 1 Answer

1. Replace the constant term $-3$ with $-3\sqrt{x^2+y^2+z^2}$. This homogenizes the inequality, and from now on you can ignore the $x^2+y^2+z^2=1$ normalization.

2. Consider substituting $a^2=x$, $b^2=y$, and $c^2=z$ so there aren't so many square roots to write out.

3. Eliminate the big nasty $\sqrt{a^4+b^4+c^4}$ by isolating it on one side and squaring, so that you end up with a 4th degree polynomial inequality. Do some simplifying. Use $\sum_{cyc}$ and $\sum_{sym}$ notation so it doesn't get too ugly.

4. I ended up with $$4abc(a+b+c) + 12\sum_{sym}a^3b \ge 22\sum_{cyc}a^2b^2.$$ From here, AM-GM will suffice; the inequality is apparently not sharp, so some terms on the left-hand side are superfluous.

• thanks! I followed through your instructive steps and it worked perfectly. – yoyostein Jul 17 '14 at 16:19