Let $a,b,c>0$. Prove: $$\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{(a+b+c)^3}{6abc(a^2+b^2+c^2)}$$

Quidditch wrote: Let $a, b, c$ be positive real numbers. Prove that $$\frac1{a^2 + bc} + \frac1{b^2 + ca} + \frac1{c^2 + ab}\leq\frac{a+b+c}{2abc}.$$ $\frac1{a^2 + bc}\le \frac1{2\sqrt{a^2bc}} =\frac{\sqrt{bc}}{2abc}\le \frac{b+c} {4abc}$

Notice that $\sum_{cyc}\frac{1}{a^2+bc}\overset{\text{T2'S}}{\ge}\frac{9}{\sum_{cyc}a^2+\sum_{cyc}ab}$ Thus we need $\frac{9}{\sum_{cyc}a^2+\sum_{cyc}ab}\ge\frac{3\sum_{cyc}ab}{2\sum_{cyc}a^2b^2}$ furthermore clearing the fractions yields $6\sum_{cyc}a^2b^2\ge\sum_{cyc}a^2\sum_{cyc}ab+\sum_{cyc}ab=\sum_{sym}a^3b+3abc\sum_{cyc}a+\sum_{cyc}a^2b^2$ Which simplifies to $5\sum_{cyc}a^2b^2\ge\sum_{sym}a^3b+3abc\sum_{cyc}a$ however notice that $(2,2,0)\succ(3,1,0)$ thus $2\sum_{cyc}a^2b^2\overset{\text{Muirhead}}{\ge}\sum_{sym}a^3b$ Therefore the inequality transforms into $3\sum_{cyc}a^2b^2\ge3abc\sum_{cyc}a\Longleftrightarrow\sum_{cyc}a^2b^2\ge abc\sum_{cyc}a=\sum_{cyc}a^2bc$ Now notice that $(2,2,0)\succ(2,1,1)$ therefore $\sum_{cyc}a^2b^2\overset{\text{Muirhead}}{\ge}\sum_{cyc}a^2bc$ so this finishes the left side of the inequality. Notice that $\frac{1}{a^2+bc}\overset{\text{AM-GM}}{\le}\frac{1}{2a\sqrt{bc}}=\frac{\sqrt{bc}}{2abc}\overset{\text{AM-GM}}{\le}\frac{b+c}{4abc}$ Therefore $\sum_{cyc}\frac{1}{a^2+bc}\le\sum_{cyc}\frac{b+c}{4abc}=\frac{1}{4}\cdot\frac{2\sum_{cyc}a}{abc}=\frac{\sum_{cyc}a}{2abc}$ so this finishes the right side of the inequaality. $$\blacksquare.$$