[Math] Find the maximum $p$ such that $2x^4y^2 + 9y^4z^2 + 12z^4x^2 – px^2y^2z^2$ is always nonnegative for all $x$, $y$, and $z$ real.

inequalityintegral-inequality

Find the maximum $p$ such that $2x^4y^2 + 9y^4z^2 + 12z^4x^2 – px^2y^2z^2$ is always nonnegative for all $x$, $y$, and $z$ real.

I have to solve this using some inequalities such as AM or GM, but I don't know how. Solutions are greatly appreciated!

Best Answer

We can simplify this into

$$2a^2b+9b^2c+12c^2a \geq pabc$$

for $a,b,c\geq 0$.Letting $a=3k,b=2m,c=n$ yields

$$36(k^2m+m^2n+n^2k) \geq (6p)kmn$$

$$k^2m+m^2n+n^2k \geq \left(\frac{p}{6}\right)kmn$$

We now use AM-GM on $k^2m, m^2n, n^2k$ to get that

$$\frac{k^2m+m^2n+n^2k}{3} \geq \left(k^3m^3n^3\right)^{\frac{1}{3}}$$

$$k^2m+m^2n+n^2k \geq 3kmn$$

So we know that this is true for all $p\leq 18$. In addition, taking $k=m=n$ makes the inequality an equality, so any $p>18$ fails. Thus we have that $p=18$ is the maximum.

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