[จูกัดเหลียง]

ข้อความเดิมเขียนโดยคุณ nooonuii

30. $a,b,c\geq 0,$ no two of them are zero

$$\dfrac{a^2}{b^2+c^2}+\dfrac{b^2}{c^2+a^2}+\dfrac{c^2}{a^2+b^2}\geq \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}$$

My_soln

WLOG $a\ge b\ge c$
by Chebyshev's
$$\sum_{cyc} \frac{a}{b+c}\cdot a(b+c)\ge\frac{1}{3}(\sum_{cyc} \frac{a}{b+c})(\sum_{cyc} a(b+c))$$
$$\Rightarrow \sum_{cyc} \frac{a}{b+c}\leq \frac{3}{2}\frac{a^2+b^2+c^2}{ab+bc+ca}$$
by Cauchy
$$\sum_{cyc} \frac{a^2}{b^2+c^2}\ge \frac{(a+b+c)^2}{2(a^2+b^2+c^2)}$$
then we need to show that
$$\frac{(a+b+c)^2}{2(a^2+b^2+c^2)}\ge \frac{3}{2}\frac{a^2+b^2+c^2}{ab+bc+ca}$$
$$\Leftrightarrow (a+b+c)^4-(a^2+b^2+c^2)((a+b+c)^2-3(a^2+b^2+c^2))\ge 0$$