If $x,y,z\ge 0$ and $\color{red}{xyz\le 1}$, show that $$\color{blue}{\dfrac{x^2-x+1}{x^2+y^2+1}+\dfrac{y^2-y+1}{y^2+z^2+1} +\dfrac{z^2-z+1}{z^2+x^2+1}\ge 1}.$$

  • $\begingroup$ Is there nice solutions? $\endgroup$
    – River Li
    Nov 28 '20 at 4:33

This inequality admits an SOS (Sum of Squares) representation \begin{align} &\frac{x^2-x+1}{x^2+y^2+1} + \frac{y^2-y+1}{y^2+z^2+1} + \frac{z^2-z+1}{z^2+x^2+1} - 1 \\ =\ & \frac{1}{(x^2+y^2+1)(y^2+z^2+1)(z^2+x^2+1)}\Big[(1-xyz)(f_1^2+f_2^2+f_3^2) + \frac{1}{12}u^TQu\Big] \end{align} where $f_1, f_2, f_3$ are polynomials with integer coefficients, $u = [1, x, y, z, xy, xz, yz, x^2, y^2, z^2, xyz, x^2y, xz^2, y^2z]^T$ and $Q$ is a positive semi-definite (PSD) $14\times 14$ constant integer matrix. The desired result follows.


1) Since the problem is tagged as contest-math but my proof is not an human proof, I do not give $f_1, f_2, f_3$ and $Q$ currently and hope to see nice proofs.

2) Mathematica Resolve can prove the inequality although we do not see the step-by-step solution.


Multiplying and dividing each of the expressions numerator by $(1 + x)$, $(1 + y)$ and $(1 + z)$ respectively:

$$ \frac{z^3+1}{(z+1) \left(x^2+z^2+1\right)}+\frac{x^3+1}{(x+1) \left(x^2+y^2+1\right)}+\frac{y^3+1}{(y+1) \left(y^2+z^2+1\right)} $$

Further simplifying:

$$ 1+ \frac{x^2-x+1}{x^2+y^2+1}-\frac{x^2+z}{x^2+z^2+1}+\frac{y^2-y+1}{y^2+z^2+1} $$

Giving similar treatment to second and fourth terms:

$$ 1+\frac{x^3+1}{(x+1) \left(x^2+y^2+1\right)}+\frac{y^3+1}{(y+1) \left(y^2+z^2+1\right)}-\frac{x^2+z}{x^2+z^2+1} $$

Note that the $1$ in the above expression suggests that this expression could be $\geq1$ as long as the sum of second, third and fourth term is $\geq0$. Simplifying further:

$$ 1+\frac{y^2-y+1}{y^2+z^2+1}-\frac{x^4+\left(y^2+z\right) x^2+x+y^2 z+z-1}{\left(x^2+y^2+1\right) \left(x^2+z^2+1\right)} $$ $$ \Rightarrow 1+\frac{y^3+1}{(y+1) \left(y^2+z^2+1\right)}+\frac{-x^4-x^2 y^2-x^2 z-x-y^2 z-z+1}{\left(x^2+y^2+1\right) \left(x^2+z^2+1\right)} $$

Note that since the second term is always positive, this means first and second terms will always be greater than 1. Now let's concentrate on third term to see if that is negative and if so, whether it is big enough to make the whole term negative.

We are given that $xyz\le1$, let us consider the upper bound of this inequality and assume $xyz=1$, this implies, $z=\frac{1}{xz}$, which upon substitute in above expression, we get:

$$ \frac{x^2+y^2+1+x y \left(-x^2+\left(2 x^3 y-x^2 y\right)+x y^3+x y-y^2-1\right)}{\left(x^2+y^2+1\right) \left(x^2 \left(x^2+1\right) y^2+1\right)} $$

How do we find out if the above term is positive or not? Let's try proof by contradiction. The denominator is always positive, so let's concentrate on the numerator and assume that the term is negative.

The approach I took was to consider only the nominator and try to solve the following equations simulataneously:

$$ 2 x^4 y^2-x^3 y^2-x^3 y+x^2 y^4+x^2 y^2+x^2-x y^3-x y+y^2+1 < 0 $$ $$x\gt0$$ $$y\gt0$$

And since there does not exist any solution so that means that numerator will be non-negative. Hence we can conclude that the term will be $\ge1$

  • $\begingroup$ I was hoping somebody will point out the mistakes, it was pretty tedious and daunting calculation. Can you tell me the mistake, I will fix it. $\endgroup$ Apr 23 '15 at 21:49
  • 1
    $\begingroup$ Please look at the difference between 4th and 2nd equation lines. When you open brackets, nominator of the second term is different, 4th degree of x shall cancel, but 3rd will emerge and other components appear. $\endgroup$
    – hOff
    Apr 24 '15 at 2:58

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