Proof of Inequality: $x^2 + xy^2 + xyz^2 \geq 4xyz - 4$

[magneto1]

Let $x,y,z$ be positive real numbers. Show that $x^2 + xy^2 + xyz^2 \geq 4xyz - 4$.

 

[Opalg]

[sp]Write the inequality as $x\Bigl(x + y\bigl(y + z(z-4)\bigr)\Bigr) \geqslant -4 \quad(*)$.

The minimum value of $z(z-4)$ is $-4$ (when $z=2$). So the left side of (*) is at least $x\bigl(x+ y(y -4)\bigr) \quad (**)$. But the minimum value of $y(y-4)$ is $-4$ (when $y=2$). So (**) is at least $x(x-4)$, which in turn is at least $-4$ (when $x=2$). This shows that (*) holds, with equality when $x=y=z=2$.[/sp]

 

[magneto1]

Excellent work. Here is another solution that uses the same idea.

Spoiler

Note that $w^2 \geq 4w - 4$ for all positive real $w$. So,
$x^2 + xy^2 + xyz^2 \geq (4x-4)+x(4y-4)+xy(4z-4)=4xyz-4$.

 

[DavidCampen]

Don't we need to show that the inequality holds as x,y,z approach zero? Has that been shown in the above?

 

[magneto1]

It is adequate to show the inequality holds for fixed and given $x,y,z \in \mathbb{R}^+$.

 

[DavidCampen]

It is adequate to show the inequality holds for fixed and given $x,y,z \in \mathbb{R}^+$.

We only needed to find a single set of values for x,y,z where the inequality holds!?

Edit:
Let z=1 then I can show that the inequality holds for an infinite number of unique tuples (x,y) where
0<x,y <= 1. The harder part is showing that it holds for all real x,y where 0<x,y,<=1.

 

[magneto1]

My apology for using the word "fixed" -- I do not mean we pin any variables to specific values.. The proofs only assume $x,y,z > 0$, and proceed to show the inequality holds for any arbitrary $x,y,z >0$ given.

 

[DavidCampen]

Thank you, my error, I thought that I had been able to find places where the inequality would not hold but I was wrong.