# An Inequality for the Cevians through Spieker Point via Brocard Angle

### Solution 1

The Spieker point can be characterized in several ways. One of these is as a point where the three triangle's cleavers intersect. Thus, in particular,

$AB+BA'=AC+CA'=\displaystyle\frac{a+b+c}{2}=s,$

i.e., $BA'=\displaystyle\frac{a+b-c}{2}\,$ and $CA'=\displaystyle\frac{a-b+c}{2}.$

Similarly we can calculate the remaining four segments. To sum up,

\displaystyle\begin{align}A'B&=B'A=\frac{a+b-c}{2},\\ A'C&=C'A=\frac{a-b+c}{2},\\ B'C&=C'B=\frac{-a+b+c}{2}. \end{align}

There is a well known expression involving the Brocard angle $\omega\,$ of $\Delta ABC:$

$\displaystyle\sin^2\omega=\frac{(-a+b+c)(a-b+c)(a+b-c)(a+b+c)}{4(a^2b^2+b^2c^2+c^2a^2)}.$

Using that and $1\gt\sin^2\omega,\,$ we obtain

\displaystyle\begin{align} 1&\gt\frac{2B'C\cdot 2C'A\cdot 2A'B\cdot 2s}{4(a^2b^2+b^2c^2+c^2a^2)}=\frac{4s\cdot B'C\cdot C'A\cdot A'B}{a^2b^2+b^2c^2+c^2a^2},\\ 1&\gt\frac{4s\cdot A'C\cdot C'B\cdot B'A}{a^2b^2+b^2c^2+c^2a^2}. \end{align}

Summing up leads to the desired result.

### Solution 2

Using Ceva's theorem, with the notations illustrated by the diagram above,

$xus'=yvt.$

Substituting products $ab, bc, ca\,$ into $\displaystyle\sum_{cycl}x^2\ge\sum_{cycl}xy,\,$ and, in the above notations, we have:

\displaystyle\begin{align} \sum_{cycl}a^2b^2 &\ge abc(a+b+c)\\ &=2s(x+y)(u+v)(s'+t)\\ &\ge 2s(2\sqrt{xy})(2\sqrt{uv})(2\sqrt{s't})\\ &=16s\sqrt{xus'}\sqrt{yvt}\\ &=16sxus'=16syvt\\ &=8s(xus'+yvt). \end{align}

It follows that

$\displaystyle\sum_{cycl}a^2b^2\ge 8s(AC'\cdot BA'\cdot CB' + AB'\cdot BC'\cdot CA').$

### Acknowledgment

The problem, with a solution (Solution 1), has been kindly communicated to me by Dan Sitaru. Solution 2 is by Soumava Chakraborty.

Soumava proof not only shows that the original estimate is off by a factor of 4, it also shows that estimate is valid for any three concurrent cevians, not just those through the Spieker point, which is, therefore, deservedly classified as a red herring.