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

[anemone]

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 $z$.

 

[anemone]

Hint:

Spoiler

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

 

[anemone]

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:

Spoiler

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 $x=y=z$.