Chebyshev Inequality toughnut Let $a^2 + b^2 + c^2 + d^2 = 1$, where $(a,b,c,d \geq 0)$. Prove that:
$$ \frac{a^{2}}{b+c+d}+\frac{b^{2}}{a+c+d}+\frac{c^{2}}{b+a+d}+\frac{d^2}{b+c+a} \geq \frac{2}{3}$$
 A: Let
$$S = \frac{a^2}{b+c+d}+\frac{b^2}{c+d+a}+\frac{c^2}{d+a+b}+\frac{d^2}{a+b+c}.$$
Using Chebyshev's inequality,
$$S \ge \frac{1}{4}\left(a^2+b^2+c^2+d^2\right)\left(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\right).$$
By the arithmetic-harmonic mean inequality,
$$\frac{1}{4}\left(\frac{1}{b+c+d}+\frac{1}{c+d+a}+\frac{1}{d+a+b}+\frac{1}{a+b+c}\right) \ge \frac{4}{3(a+b+c+d)}.$$
Lastly, by the arithmetic-quadratic mean inequality,
$$a+b+c+d \le 4 \sqrt{\frac{a^2+b^2+c^2+d^2}{4}}=2.$$
Therefore,
$$S \ge \frac{4}{3(a+b+c+d)} \ge \frac{4}{6}=\frac{2}{3}.$$
A: I think C-S is better here:
$$\sum_{cyc}\frac{a^2}{b+c+d}=\sum_{cyc}\frac{a^4}{a^2(b+c+d)}\geq\frac{(a^2+b^2+c^2+d^2)^2}{\sum\limits_{cyc}a^2(b+c+d)}=\frac{1}{\sum\limits_{cyc}a(b^2+c^2+d^2)}.$$
Thus, it remains to prove that
$$3\geq2\sum_{cyc}a(1-a^2)$$ or
$$\sum_{cyc}\left(2a^3-2a+\frac{3}{4}\right)\geq0$$ or
$$\sum_{cyc}(2a-1)(4a^2+2a-3)\geq0$$ or
$$\sum_{cyc}\left((2a-1)(4a^2+2a-3)+\frac{1}{2}(4a^2-1)\right)\geq0$$ or
$$\sum_{cyc}(2a-1)^2(4a+5)\geq0.$$
Done!
