<s>[INST] 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.
$$12(ab+ba+ac) < 7a^2+15b^2+18c^2$$
$$(2a-3b)^2+(2b-3c)^2+(2a-3c)^2+2b^2-a^2>0$$
 [/INST] 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:
$$\frac{5}{2}a^2+\frac{13}{2}b^2-8ab\geq2\sqrt{\frac{5}{2}a^2\cdot\frac{13}{2}b^2}-8ab=\left(\sqrt{65}-8\right)ab>0$$
</s>