Prove Inequality: $x^2y\,+\,y^2z\,+\,z^2x \ge 2(x\,+\,y\,+\,z) - 3$

[anemone]

Given that $x,\,y$ and $z$ are positive real numbers such that $xy + yz + zx = 3xyz.$

Prove that $x^2y + y^2z + z^2x\ge 2(x + y + z) − 3$.

 

[Euge]

My solution:

Spoiler

Dividing the equation $xy + yz + zx = 3xyz$ by $xyz$ yields $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3$. So

$$x^2y + y^2z + z^2x + 3 = \left(x^2y + \frac{1}{y}\right) + \left(y^2z + \frac{1}{z}\right) + \left(z^2x + \frac{1}{x}\right)$$
$$ \ge 2\sqrt{x^2y/y} + 2\sqrt{y^2z/z} + 2\sqrt{z^2x/x} = 2x + 2y + 2z = 2(x + y + z).$$

Therefore

$$x^2y + y^2z + z^2x \ge 2(x + y + z) - 3.$$

 

[anemone]

My solution:

Spoiler

Dividing the equation $xy + yz + zx = 3xyz$ by $xyz$ yields $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 3$. So

$$x^2y + y^2z + z^2x + 3 = \left(x^2y + \frac{1}{y}\right) + \left(y^2z + \frac{1}{z}\right) + \left(z^2x + \frac{1}{x}\right)$$
$$ \ge 2\sqrt{x^2y/y} + 2\sqrt{y^2z/z} + 2\sqrt{z^2x/x} = 2x + 2y + 2z = 2(x + y + z).$$

Therefore

$$x^2y + y^2z + z^2x \ge 2(x + y + z) - 3.$$

Very well done, Euge! (Cool)

My solution:

Spoiler

\(\displaystyle \begin{align*}x^2y + y^2z + z^2x&=\frac{x^2}{\frac{1}{y}}+\frac{y^2}{\frac{1}{z}}+\frac{z^2}{\frac{1}{x}}\\&\ge \frac{(x+y+z)^2}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}\text{by the Titu's Lemma}\\&=\frac{(x+y+z)^2}{3}\text{since we're told $xy + yz + zx = 3xyz.$}\end{align*}\)

Now if we can prove \(\displaystyle \frac{(x+y+z)^2}{3}\ge 2(x + y + z) − 3\) then we're done.

But that is true since

\(\displaystyle \frac{(x+y+z)^2}{3}\ge 2(x + y + z) − 3\)

\(\displaystyle (x+y+z)^2- 6(x + y + z)+ 9\ge 0\)

\(\displaystyle (x+y+z-3)^2\ge 0\) for all positive real $x,\,y$ and $z$.