Finding the infimum of a multivariable polynomial We are given a polynomial $P(x,y,z) = x^4y^2+x^2y^4+z^6-3x^2y^2z^2$ we need to prove that $\text {inf}_{x,y,z}p(x,y,z) =0 $.
I tried to prove it using the property that a point will be minimum if $\nabla f = 0$ and $\text{H}(x)\ge0$.
$$\nabla f = \begin{bmatrix} 
 4x^3+2xy^4-6xy^2z^2\\
 2x^4y+4x^2y^3-6x^2yz^2\\
 6z^5-6x^2y^2z\\
 \end{bmatrix}  $$ gives $x=0,y=\text{k},z=0$ and we observe that $\text{H}(x)=0$ thus it has minimum at $(0,\text{k},0)$ and thus $\text {inf}_{x,y,z}p(x,y,z) =0 $.
I have assumed that the minimum point will also give the infimum point.
Is my understanding correct?
Any suggestions or hint will be very helpful, thanks in advance!
 A: Proving that $f$ has a local minimum at $(0,0,0)$ doesn't prove that $f(0,0,0)=0$ is a global mimimum value for $f$.

How do you know the range of $f$ has no negative values?

Moreover, you haven't actually proved that $f$ has a local minimum at $(0,0,0)$, since at the critical point $(0,0,0)$, the Hessian is the zero matrix, which is inconclusive as to the type of critical point.

But let's suppose you somehow managed to prove that $f$ has a local minimum at $(0,0,0)$. Then your idea (as you suggested in the comments) was to show that $f$ is convex, in which case, a local minimum would also be a global minimum.

But in fact, $f$ is not convex, since for example, using the points
$$
A=(1,1,1),\;\;\;\;B=(1,1,-1)
$$
we have
$$
f\left(\frac{A+B}{2}\right)=f(1,1,0)=2
$$
whereas
$$
\frac{f(A)+f(B)}{2}
=
\frac{f(1,1,1)+f(1,1,-1)}{2}
=
\frac{0+0}{2}
=
0
$$
In any case, you missed a much easier approach . . .

By the AM-GM inequality, for all real $x,y,z$, we have
$$
\frac{x^4y^2+x^2y^4+z^6}{3}\ge\sqrt[3]{x^6y^6z^6}=x^2y^2z^2
$$
hence $f(x,y,z)\ge 0$.

Then since $f(0,0,0)=0$, it follows that $0$ is a global mimimum value for $f$.
