Question 3
For positive real numbers $a,b,c$, prove
$$\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a}\ge\frac23(a^2+b^2+c^2).$$
Proposed by Titu Andreescu (credit in the supplied notes).
$$\frac{a^3+3b^3}{5a+b}+\frac{b^3+3c^3}{5b+c}+\frac{c^3+3a^3}{5c+a}\ge\frac23(a^2+b^2+c^2).$$
Proposed by Titu Andreescu (credit in the supplied notes).