<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] 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.
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