MHB What Is the Maximum Value of \(a\) in This Polynomial Inequality?

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The discussion focuses on determining the maximum value of \(a\) in the polynomial inequality \(x^4+y^4+z^4+xyz(x+y+z) \geq a(xy+yz+zx)^2\). Participants are encouraged to explore various approaches to find the greatest value of \(a\) that satisfies the inequality for all real numbers \(x\), \(y\), and \(z\). The hint emphasizes the importance of testing different values and configurations of \(x\), \(y\), and \(z\) to derive the solution. The conversation may include algebraic manipulations and potential substitutions to simplify the problem. Ultimately, the goal is to establish the conditions under which the inequality holds true for all specified variables.
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Find the greatest value of $a$ for which the inequality $x^4+y^4+z^4+xyz(x+y+z)≥ a(xy+yz+zx)^2$ holds for all values ​​of $x ,\,y$ and $z$.
 
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Hint:

$x^2+y^2+z^2\ge xy+yz+xz$
 
anemone said:
Find the greatest value of $a$ for which the inequality $x^4+y^4+z^4+xyz(x+y+z)≥ a(xy+yz+zx)^2$ holds for all values ​​of $x ,\,y$ and $z$.

My solution:

Note that

$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2(x^2y^2+y^2z^2+x^2z^2)$ and $(xy+yz+xz)^2=x^2y^2+y^2z^2+x^2z^2+2xyz(x+y+z)$ so the LHS of the inequality can be rewritten as:

$\begin{align*}x^4+y^4+z^4+xyz(x+y+z)&=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+x^2z^2)+\dfrac{(xy+yz+xz)^2}{2}-\dfrac{x^2y^2+y^2z^2+x^2z^2}{2}\\&=(x^2+y^2+z^2)^2-5\left(\dfrac{x^2y^2+y^2z^2+x^2z^2}{2}\right)+\dfrac{(xy+yz+xz)^2}{2}\end{align*}$

From Cauchy Schwarz inequality we have

$(x^2+y^2+z^2)^2\ge 3(x^2y^2+y^2z^2+x^2z^2)\ge (xy+yz+xz)^2$

Therefore we get

$\begin{align*}x^4+y^4+z^4+xyz(x+y+z)&=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+x^2z^2)+\dfrac{(xy+yz+xz)^2}{2}-\dfrac{x^2y^2+y^2z^2+x^2z^2}{2}\\&=(x^2+y^2+z^2)^2-5\left(\dfrac{x^2y^2+y^2z^2+x^2z^2}{2}\right)+\dfrac{(xy+yz+xz)^2}{2}\\&\ge (xy+yz+xz)^2-5\left(\dfrac{\dfrac{(xy+yz+xz)^2}{3}}{2}\right)+\dfrac{(xy+yz+xz)^2}{2}\\&\ge \dfrac{2}{3}(xy+yz+zx)^2\end{align*}$

Therefore the greatest value of $a$ is $\dfrac{2}{3}$, equality occurs at $x=y=z$.
 
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