Team Selection Test for JBMO

Let $f(a,b,c)=\frac{a^4+5b^4}{a(a+2b)}+\frac{b^4+5c^4}{b(b+2c)}+\frac{c^4+5a^4}{c(c+2a)}$. Since $a+b+c=1$ we will prove that $f(a,b,c)\ge a^2+b^2+c^2+ab+bc+ac$.

By Cauchy-Schwarz inequality for positive $x_1,\dots ,x_n$ $$(x_1+\cdots +x_n)\left(\frac{a_1^2}{x_1}+\cdots +\frac{a_n^2}{x_n}\right)\ge (a_1+\cdots +a_n{)}^2(1)$$ By applying (1) we get $$\frac{a^4}{a(a+2b)}+\frac{b^4}{b(b+2c)}+\frac{c^4}{c(c+2a)}\ge \frac{(a^2+b^2+c^2{)}^2}{(a+b+c{)}^2}=(a^2+b^2+c^2{)}^2(2)$$ $$\frac{b^4}{a(a+2b)}+\frac{c^4}{b(b+2c)}+\frac{a^4}{c(c+2a)}\ge \frac{(a^2+b^2+c^2{)}^2}{(a+b+c{)}^2}=(a^2+b^2+c^2{)}^2(3)$$ Now (2) + 5 (3) gives $f(a,b,c)\ge 6(a^2+b^2+c^2{)}^2$. Thus, the proof of $$6(a^2+b^2+c^2{)}^2\ge a^2+b^2+c^2+ab+bc+ac(4)$$ will complete the solution. Now the sum of $\frac{a^2+b^2}{2}\ge ab$, $\frac{b^2+c^2}{2}\ge bc$, $\frac{c^2+a^2}{2}\ge ca$ gives $a^2+b^2+c^2\ge ab+bc+ca$. Therefore, (4) will follow from $6(a^2+b^2+c^2{)}^2\ge 2(a^2+b^2+c^2)$ or $3(a^2+b^2+c^2{)}^2\ge 1$. But since $a+b+c=1$ the last inequality is a quadratic-arithmetic means inequality. Done.