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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.

Oshawott
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    For what it is worth, Wikipedia calls this the Chebyshev sum inequality, to distinguish it from a probability inequality also bearing Chebyshev’s name. https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality – Thomas Andrews Sep 13 '21 at 16:48
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    The part in red seems to be wrong: Try $(x, y, z) = (3, 2, 1)$. – Martin R Sep 13 '21 at 16:51
  • Or $(x, y, z) = (2, 1, 1)$. Perhaps the inequality holds the other way around, but that proof is simply wrong. – Martin R Sep 13 '21 at 16:59
  • Yep, it’s wrong. – Thomas Andrews Sep 13 '21 at 16:59
  • @MartinR I also think that. But the solution is from a book and also appears in this AoPS thread. That's why I'm confused. – Oshawott Sep 13 '21 at 17:03
  • What answer do you expect? The inequality $x^3y^2 + y^3z^2 + z^3y^2 \ge \frac 13 (x^3+y^3+z^3)(x^2+y^2+z^2)$ does not hold in general, there are numerous counterexamples. That is a wrong proof. – Martin R Sep 13 '21 at 17:06
  • Well, since the step is clearly wrong for specific values of $(x,y,z),$ it is wrong no matter how many sources agree. AoPS likely used the same source for the solution. – Thomas Andrews Sep 13 '21 at 17:06
  • Note that this inequality has been solved in several threads already: https://approach0.xyz/search/?q=OR%20content%3A%24%5Cfrac%20%7Bx%5E%7B2%7Dy%7D%7Bz%7D%20%2B%20%5Cfrac%20%7By%5E%7B2%7Dz%7D%7Bx%7D%20%2B%20%5Cfrac%20%7Bz%5E%7B2%7Dx%7D%7By%7D%5Cgeq%20x%5E%7B2%7D%20%2B%20y%5E%7B2%7D%20%2B%20z%5E%7B2%7D%24%2C%20AND%20site%3Amath.stackexchange.com – Martin R Sep 13 '21 at 17:13
  • @MartinR I know but I didn't find one that uses Chebyshev inequality or rearrangement inequality. But I intuitively believe that those can be used in this problem. That's why I'm asking. – Oshawott Sep 13 '21 at 17:17
  • @Unknown See my solution here: https://math.stackexchange.com/questions/555400 – Michael Rozenberg Sep 13 '21 at 20:09
  • As @ThomasAndrews pointed out, I think you should edit your question and not refer to "Chebyshev's sum inequality" as "Chebyshev's inequality", as it will lead to confusion (or at least be more explicit about the difference). – econbernardo Sep 15 '21 at 13:01
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    @econbernardo Thanks, I edited my question. – Oshawott Sep 15 '21 at 13:07

1 Answers1

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(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 $z = 0$.

marty cohen
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