Bad Ineq

We simplify $\dfrac{a^2}{bc} + \dfrac{b^2}{ac} + \dfrac{c^2}{ab} + 3(\dfrac{a+b}{c} + \dfrac{a+c}{b} + \dfrac{b+c}{a} )$ to $\dfrac{a^3 +3a^2b + 3a^2c + b^3 + 3ab^2 + 3b^2c + c^2 + ac^2 + bc^2}{abc}$ , which factors to $\dfrac{(a+b+c)^3 - 6abc}{abc}$ . Let this be equal to $x$ . Now, multiply by $abc$ to get $(a+b+c)^3 - 6abc = abcx$ . Add $6abc$ to both sides to get $a+b+c)^3 = abc(x+6)$ . Now, divide by $x+6$ to get $\dfrac{(a+b+c)^3}{x+6} = abc$ . Take the cubic root to get $\dfrac{a+b+c}{\sqrt[3]{x+6}} = \sqrt[3]{abc}$ . We know by AM-GM that $\dfrac{a+b+c}{3} \ge \sqrt[3]{abc}$ . If $\sqrt[3]{x+6} < 3$ , then we will have $\dfrac{a+b+c}{\sqrt[3]{x+6}} < \sqrt[3]{abc}$ , which is impossible. Thus, $\sqrt[3]{x+6} \ge 3 \rightarrow x+6 \ge 27 \rightarrow x \ge 21$ and the minimum value is $\boxed{21}$ .

When we simplify the numerator we get,

{a³+b³+c³+3ab(a+b)+3ac(a+c)+3bc(b+c)} /abc

={(a+b+c)³-6abc}/abc

Now by AM≥GM

(a+b+c/3)≥³√abc

(a+b+c)³/27≥abc

(a+b+c)³/abc≥27

So our original expression ≥27-6=21