Find the largest constant $\mathrm{C}$ such that for all $x, y, z \geq 0,(y z+z x+$ $x y)^{2}(x+y+z) \geq C x y z\left(x^{2}+y^{2}+z^{2}\right)$.
By setting $y=z=1, \frac{(y z+z x+x y)^{2}(x+y+z)}{x y z\left(x^{2}+y^{2}+z^{2}\right)}=\frac{(2 x+1)^{2}(x+2)}{x\left(x^{2}+2\right)}$. This quantity tends to 4 as $x$ tends to $+\infty$, so we will try to prove the inequality with $C=4$.

By expanding, we need to show that $\left(x^{3} y^{2}+x^{2} y^{3}+y^{3} z^{2}+y^{2} z^{3}+z^{3} x^{2}+z^{2} x^{3}\right)+5\left(x^{2} y^{2} z+y^{2} z^{2} x+z^{2} x^{2} y\right) \geq 2\left(x^{3} y z+y^{3} z x+z^{3} x y\right)$.

We then draw the triangle:

![](https://cdn.mathpix.com/cropped/2024_05_10_da6e2ad19b1b4e3d33dcg-205.jpg?height=621&width=632&top_left_y=611&top_left_x=752)

And we see that it suffices to apply the AM-GM inequality three times with the extreme black points $\left(x^{2} z^{3}+z^{3} y^{2} \geq 2 z^{3} x y\right.$, etc.) to conclude.