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

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.

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

• Please show your work on completing the square. Oct 27, 2020 at 19:39
• I am inclined to say that this should hold for all real numbers, actually. In general, changing some of the variables from positive to negative should keep the right side the same but should only ever have the potential to lower the left side, so if this inequality is true for positive numbers, it should hold over all reals too. Oct 27, 2020 at 19:44
• @boink all non-zero, you mean? Oct 27, 2020 at 19:48
• @ParclyTaxel No, all positive Oct 27, 2020 at 19:50

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.

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 $$f(a,b,c)>0$$.

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: $$\frac{5}{2}a^2+\frac{13}{2}b^2-8ab\geq2\sqrt{\frac{5}{2}a^2\cdot\frac{13}{2}b^2}-8ab=\left(\sqrt{65}-8\right)ab>0$$