Let $x\geq y\geq z>0$. Prove that $ \frac {x^{2}y}{z} + \frac {y^{2}z}{x} + \frac {z^{2}x}{y}\geq x^{2} + y^{2} + z^{2}.$

Question

Let $x\geq y\geq z>0$. Prove that $$ \frac {x^{2}y}{z} + \frac {y^{2}z}{x} + \frac {z^{2}x}{y}\geq x^{2} + y^{2} + z^{2}.$$

The problem is from Vietnamese MO 1991 and has been posted here before. However I am having some issues on a particular solution (that is not on the given link). The solution is from the book "Inequalities from around the world (1990-2005)" which is as follows:

Since $x\geq y\geq z>0$ we have that $$\begin{align} \frac {x^2y}{z} + \frac {y^2z}{x} + \frac {z^2x}{y} &= \frac {\color\red{x^3y^2 + y^3z^2 + z^3x^2}}{xyz}\\ &\geq \frac {\color\red{(x^3 + y^3 + z^3)(x^2 + y^2 + z^2)}}{\color\red{3}(xyz)}\\ &\geq \frac {3xyz(x^2 + y^2 + z^2)}{3(xyz)}\\ &= (x^2 + y^2 + z^2) \end{align}$$ by Chebyshev sum inequality.$\ \ \ \blacksquare$

The solution seem to use Chebyshev sum inequality on $(x^3,y^3,z^3)$ and $(y^2,z^2,x^2)$ [the part in red]. But the problem is that the first sequence is decreasing while the second is not. To use Chebyshev inequality, the second sequence must be $(x^2,y^2,z^2)$ and then the result will be $$\begin{align} x^5+y^5+z^5\geq \frac{(x^3 + y^3 + z^3)(x^2 + y^2 + z^2)}{3} \end{align}$$ which is not helpful in the problem.

So, is the solution wrong or am I missing something?

Edit: From the comments, I realized that the solution is wrong. So can this problem be solved with Chebyshev sum inequality or rearrangement inequality? I have some solutions but it seems to me that the problem can be solved with those inequalities but I didn't find one.

Answer 1

(Note: This was done without looking at any of the links. It probably duplicates some of them.)

Let $y = (1+b)z, x = (1+a)y, a, b \ge 0$.

$x = (1+a)(1+b)z$.

$\begin{array}\\ RHS &=z^2((1+a)^2(1+b)^2+(1+b)^2+1)\\ LHS &=\dfrac {x^{2}y}{z} + \dfrac {y^{2}z}{x} + \dfrac {z^{2}x}{y}\\ &=\dfrac {((1+a)(1+b)z)^{2}(1+b)z}{z} + \dfrac {((1+b)z)^{2}z}{(1+a)(1+b)z} + \dfrac {z^{2}(1+a)(1+b)z}{(1+b)z}\\ &=(1+a)^2(1+b)^3z^2 + \dfrac {(1+b)z^2}{1+a} + z^{2}(1+a)\\ &=z^2\left((1+a)^2(1+b)^3 + \dfrac {1+b}{1+a} + (1+a)\right)\\ \end{array} $

so we want

$(1+a)^2(1+b)^3 + \dfrac {1+b}{1+a} + (1+a) \ge ((1+a)^2(1+b)^2+(1+b)^2+1)$.

Throwing this into Wolfy (this could be done by hand),

$(1+a)^2(1+b)^3 + \dfrac {1+b}{1+a} + (1+a) -((1+a)^2(1+b)^2+(1+b)^2+1)\\ =(2 a^2 + 4 a + 1) b^2 + \dfrac{a (a^2 + 3 a + 1) b}{a + 1} + \dfrac{a^2}{a + 1} + (a + 1)^2 b^3\\ \ge 0 $.

There is equality if and only if $a=b=0$ or $z = 0$.