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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!

indmath
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  • It could also happen that as some of $x,y,z$ go to $\infty$ and the others to $-\infty$, the function blows up or goes below $0$. This can't be checked with derivatives, so you will need better tools : but then you may as well use AM-GM, for example. – Sarvesh Ravichandran Iyer Sep 02 '21 at 03:12
  • @TeresaLisbon If we show that f is convex then can we comment that the critical point will give the global minimum? Thanks! – indmath Sep 02 '21 at 03:22
  • @indmath If you show convexity then you are good : any local minimum must be a global minimum as well, for a convex function. So getting one local minimum will be enough. But do make sure your function is convex. – Sarvesh Ravichandran Iyer Sep 02 '21 at 03:24
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    It may be a sum of squares (of homogeneous polynomials). Compare https://math.stackexchange.com/questions/4064630/choi-lam-homogeneous-polynomials-as-sums-of-squares – Will Jagy Sep 02 '21 at 03:26
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    This is the homogeneous version of the Motzkin polynomial. It is not the sum of squares of polynomials. However, multiply it by $x^2 + y^2 + z^2$ and the result is S.O.S. – Will Jagy Sep 02 '21 at 03:43

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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 $f$.

quasi
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  • Thanks for the solution! Suppose if we want to prove it without the AM-GM inequality can we try to show that if f is a convex function than the critical point will also be the global minimum? – indmath Sep 02 '21 at 03:20
  • Yes agreed, trying that. Thanks! – indmath Sep 02 '21 at 03:41
  • @indmath "the critical point" $;-;$ You shouldn't call it *the* critical point since it's certainly not unique. The minimum is attained for $,x=z=0 ;\lor; y=z=0 ;\lor; \pm x = \pm y = \pm z,$. – dxiv Sep 02 '21 at 03:42
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    This is the homogeneous version of the Motzkin polynomial. It is not the sum of squares of polynomials. However, multiply it by $x^2 + y^2 + z^2$ and the result is S.O.S. – Will Jagy Sep 02 '21 at 03:43