# A Not Quite Cyclic Inequality from Tibet

### Solution

Since the required inequality is homogeneous, we may specify $x+y+z=3$ and then $xy+yz+zx=3((1-t),$ $0\le t \lt 1$ and $x^2+y^2+z^2=3(1+2t).$

We need to prove that

$(3-z)^2(2+t)^2\ge 8(1-t)^2[3(1+2t)-z^2]$

which is equivalent to $az^2+bz+c\ge 0$ where

\begin{align} a&=(2+t)^2+8(1-t)^2,\\ b&=-6(2+t)^2,\\ c&=3[3(2+t)^2-8(1-t)^2(1+2t)]. \end{align}

For the discriminant $\Delta$ of this quadratic equation we have

\displaystyle\begin{align}-\frac{\Delta}{192(1-t)^2}&=(2+t)^2-4(1-t)^2(1+2t)\\ &=t(13t+4-8t^2)\ge 0 \end{align}

for $t\in [0,1).$

### Acknowledgment

The following problem, due to Liu-Bao Qian, Lhasa Information Center, Tibet, was kindly communicated to me by Leo Giugiuc, along with a solution of his.

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Copyright © 1996-2018 Alexander Bogomolny
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