**Problem**: (2000 USA TST - #1) Given three non-negative reals $a,b,c$ prove that

$\frac{a+b+c}{3}-\sqrt[3]{abc} \le \max\{(\sqrt{a}-\sqrt{b})^2, (\sqrt{b}-\sqrt{c})^2, (\sqrt{c}-\sqrt{a})^2\}$.

**Solution**: Let's prove a stronger inequality, that

$a+b+c-3(\sqrt[3]{abc}) \le (\sqrt{a}-\sqrt{b})^2+(\sqrt{b}-\sqrt{c})^2+ (\sqrt{c}-\sqrt{a})^2$.

Make sure you understand why this is stronger (notice that the new RHS is less than three times the old RHS). So we have some ugly radicals here that we'd like to get rid of - try substituting $a = x^6, b = y^6, c = z^6$. Our inequality becomes

$x^6+y^6+z^6-3x^2y^2z^2 \le (x^3-y^3)^2+(y^3-z^3)^2+(z^3-x^3)^2$

which reduces down to

$2(x^3y^3+y^3z^3+z^3x^3) \le x^6+y^6+z^6+3x^2y^2z^2$.

But by Schur, we know

$\displaystyle \sum_{cyc} x^2(x^2-y^2)(x^2-z^2) \ge 0$

$ \displaystyle x^6+y^6+z^6 +3x^2y^2z^2 \ge \sum_{cyc} (x^4y^2+x^4z^2) $,

where a cyclic summation is taken over the variables "cycled" in order (e.g. $\displaystyle \sum_{cyc} x^2y = x^2y+y^2z+z^2x$).

Then, by AM-GM, we have $ \displaystyle \sum_{cyc} (x^4y^2+x^4z^2) = \sum_{cyc} (x^4y^2+y^4x^2) \ge \sum_{cyc} (2x^3y^3)$, so

$ \displaystyle x^6+y^6+z^6 +3x^2y^2z^2 \ge \sum_{cyc} (2x^3y^3) = 2(x^3y^3+y^3z^3+z^3x^3) $

as desired. QED.

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Comment: This is pretty easy for a USA TST problem, even if it was a #1. The strongest inequality necessary was Schur and it was already all homogenized and everything.

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Practice Problem: Prove the inequality by assuming WLOG $(\sqrt{a}-\sqrt{b})^2$ is the maximum of the three.

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