Prove that $12(ab+ba+ac) <7a^2+15b^2+18c^2$ holds for all positive numbers.

Question

Prove that $12(ab+ba+ac) < 7a^2+15b^2+18c^2$ holds for all positive numbers.

I tried completing the square, but that solution would suggest that inequality holds for all real numbers. Inequalities between means did not work for me either.

$$12(ab+ba+ac) < 7a^2+15b^2+18c^2$$ $$(2a-3b)^2+(2b-3c)^2+(2a-3c)^2+2b^2-a^2>0$$

Answer 1

By AM-GM, or completing the square,

$(p-q)^2 \ge 0 \Rightarrow p^2 + q^2 \ge 2pq$

we have

$$\color{blue}{(4a^2 + 9b^2)} + \color{red}{(3a^2 + 12c^2)} + \color{green}{(6b^2 + 6c^2)} \ge \color{blue}{12ab} + \color{red}{12ca} + \color{green}{12bc}$$

with equality for $2a=3b$, $a=2c$, $b=c$ whose simultaneous solution is $(a,b,c)=(0,0,0)$

For $a,b,c > 0$, we have strict inequality.

Answer 2

We write $f(a,b,c)=7a^2+15b^2+18c^2-12(ab+bc+ca)$ as a quadratic form with matrix $$\begin{bmatrix} 7&-6&-6\\ -6&15&-6\\ -6&-6&18\end{bmatrix}$$ This can be verified to be a positive-definite matrix, so $f(a,b,c)\ge0$ for all $a,b,c\in\mathbb R$. In particular, if $a,b,c>0$ then $f(a,b,c)>0$.

Answer 3

The completing to square works:

We need to prove that: $$18c^2-12(a+b)c+7a^2+15b^2-12ab>0$$ or $$9c^2-6(a+b)c+\frac{7}{2}a^2+\frac{15}{2}b^2-6ab>0$$ or $$(3c-a-b)^2+\frac{5}{2}a^2+\frac{13}{2}b^2-8ab>0,$$ which is true by AM-GM: $$\frac{5}{2}a^2+\frac{13}{2}b^2-8ab\geq2\sqrt{\frac{5}{2}a^2\cdot\frac{13}{2}b^2}-8ab=\left(\sqrt{65}-8\right)ab>0$$