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

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SUMMARY

The maximum value of \(a\) in the polynomial inequality \(x^4+y^4+z^4+xyz(x+y+z) \geq a(xy+yz+zx)^2\) is determined to be 1. This conclusion is reached through the application of symmetric polynomial techniques and the AM-GM inequality. The inequality holds for all real values of \(x\), \(y\), and \(z\) when \(a\) is set to 1, ensuring the left-hand side remains greater than or equal to the right-hand side across all valid inputs.

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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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