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I'm trying to determine if $\bigl(x+y)^4(y+z)^4(z+x)^4 \geq$ $8x^2y^2z^2\bigl((x+y)^2 + (y+z)^2\bigr)\bigl((y+z)^2 + (z+x) ^2\bigr)\bigl((z+x)^2 + (x+y)^2\bigr)$

for $x,y,z>0$.

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    I edited my expression to be more clear.2012-08-06

3 Answers 3

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One way to attack it is to compute a sum-of-squares solution numerically through semidefinite programming, i.e., finding a positive semidefinite matrix $Q$ such that your polynomial $p(x,y,z)$ can be written as $v^TQv$ where $v$ is a suitably selected vector of monomials in $x,y,z$.

The following piece of code computes a sum-of-squares solution in MATLAB by using the toolbox YALMIP. It assumes that you have an efficient semidefinite solver installed, such as SeDuMi or SDPT3

sdpvar s t u x = s^2; y = t^2; z = u^2; p = (x+y)^4*(y+z)^4*(z+x)^4-8*x^2*y^2*z^2*((x+y)^2+(y+z)^2)*((y+z)^2+(x+z)^2)*((z+x)^2+(x+y)^2); [diags,v,Q] = solvesos(sos(p)); 

The solution $Q$ is a 61x61 matrix, close to singular, positive semidefinite matrix (with some block-structure arising due to symmetry exploitation performed in the decomposition). Note though, it is only a numerical indication of positivity. The result is only correct up to roughly 7 or 8 digits, since there is a discrepancy between p and the computed decomposition.

max(abs(coefficients(p-v{1}'*Q{1}*v{1},[s;t;u]))) 

However, although not a true certificate, it might give some hints towards a symbolic decomposition, and hints at what the answer is.

By attacking @Camerons polynomials instead, numerics are more to our advantage

sdpvar s t a = 2 + s^2; b = 2 + t^2; p = ((a+b-2)*a*b)^4-8*(a-1)*(b-1)^2*((a+b-2)^2+a^2)*((a+b-2)^2+b^2)*(a^2+b^2) [diagnostics,v,Q] = solvesos(sos(p)) 

This decomposition has a 68x68 matrix Q with smallest eigenvalue ~0.45, and by using some trivial results relating the difference between $p$ and $v^TQv$ with the smallest eigenvalue of $Q$, it can be shown that this decomposition actually proves positivity, despite being approximate(Theorem 4). Alas, everything is still only correct to machine precision etc, i.e., still not a truly valid certificate.

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Updated Version: Call the given polynomial $f(x,y,z)$. Note that this is a symmetric polynomial--that is, if we permute $x,y,z$ in the polynomial, we end up with the same polynomial--so without loss of generality, we may assume that $x,y\geq z$.

Another nice observation is that every term of the expanded polynomial will have the same total degree--namely $12$--so the polynomial is homogeneous, too. This suggests that we might want to rewrite $x=\alpha z$, $y=\beta z$, where $\alpha,\beta\geq 1$ (since $x,y\geq z>0$), for then each term will have a factor of $z^{12}$, and our $3$-variable problem reduces to the (less daunting) $2$-variable problem of determining whether $\frac{f(\alpha z,\beta z,z)}{z^{12}}\geq 0$ for all $\alpha,\beta\geq 1$.

To make our lives slightly easier (so we aren't working with so much at once and everything fits on one line), let's set $p(x,y,z):=\bigl((x+y)(x+z)(y+z)\bigr)^4$ and $q(x,y,z):=8x^2y^2z^2\bigl((x+y)^2+(x+z)^2\bigr)\bigl((x+y)^2+(y+z)^2\bigr)\bigl((x+z)^2+(y+z)^2\bigr),$ so that $f=p-q$. Now, $p(\alpha z,\beta z,z)=\bigl((\alpha+\beta)(\alpha+1)(\beta+1)\bigr)^4z^{12}$ and $q(\alpha z,\beta z,z)=8\alpha^2\beta^2\bigl((\alpha+\beta)^2+(\alpha+1)^2\bigr)\bigl((\alpha+\beta)^2+(\beta+1)^2\bigr)\bigl((\alpha+1)^2+(\beta+1)^2\bigr)z^{12}.$

These are still messy, so let's perform another substitution for increased simplicity. Set $a:=\alpha+1$, $b:=\beta+1$, so $p(\alpha z,\beta z,z)=\bigl((a+b-2)ab\bigr)^4z^{12}$ and $q(\alpha z,\beta z,z)=8(a-1)^2(b-1)^2\bigl((a+b-2)^2+a^2\bigr)\bigl((a+b-2)^2+b^2\bigr)(a^2+b^2)z^{12}.$ Hence, we need to show that $\bigl((a+b-2)ab\bigr)^4-8(a-1)^2(b-1)^2\bigl((a+b-2)^2+a^2\bigr)\bigl((a+b-2)^2+b^2\bigr)(a^2+b^2)\geq 0$ for all $a,b\geq 2$.

At this point, though, I'm stuck. Perhaps this will give you (or someone else) an idea on how to get the rest of the way--or perhaps Mathematica will give you what you need. Sorry!

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    I edited just before to determine not >0 but $\ge0$ .. Sorry for my first mistake.2012-08-06
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The first proof (with calculator).

Let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+u$.

Hence, $u\geq0$, $v\geq0$ and $\prod_{cyc}(x+y)^4-8x^2y^2z^2\prod_{cyc}((x+y)^2+(x+z)^2)=$ $=1024(u^2-uv+v^2)x^{10}+1280(2u^3+u^2v+uv^2+2v^3)x^9+$ $+192(13u^4+34u^3v+39u^2v^2+34uv^3+13v^4)x^8+$ $+32(38u^5+217u^4v+518u^3v^2+518u^2v^3+217uv^4+38v^5)x^7+$ $+8(38u^6+416u^5v+1853u^4v^2+3092u^3v^3+1853u^2v^4+416uv^5+38v^6)x^6+$ $+16(2u^7+50u^6v+415u^5v^2+1143u^4v^3+1145u^3v^4+415u^2v^5+50uv^6+2u^7)x^5+$ $+8(10u^6+199u^5v+898u^4v^2+1416u^3v^3+898u^2v^4+199uv^5+10v^6)uvx^4+$ $+16(12u^5+97u^4v+229u^3v^2+229u^2v^3+97uv^4+12v^5)u^2v^2x^3+$ $+8(u^6+22u^5v+77u^4v^2+114u^3v^3+77u^2v^4+22uv^5+v^6)u^2v^2x^2+$ $+8(u+v)^3(u^2+3uv+v^2)u^3v^3x+(u+v)^4u^4v^4\geq0.$ Done!

The second way (by hand) it's $uvw$ https://www.artofproblemsolving.com/community/c6h278791

Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.

Hence, our inequality it's $f(w^3)\geq0$, where $f$ is a concave function.

I am ready to explain the last statement if you want.

Hence, it's enough to prove our inequality for an extremal value of $w^3$,

which happens in the following cases.

  1. $w^3\rightarrow0^+$.

This case is obvious;

  1. $y=z=1$, which gives $(x-1)^2(x+1)^2(2x^3+5x^2+8x+1)\geq0.$ Done again!