An inequality with Three Points

Solution

Let $P\,$ be a point in the interior of $\Delta ABC.\,$ Then, obviously,

$\displaystyle \frac{[\Delta APB]+[\Delta BPC]+[\Delta CPA]}{[\Delta ABC]}=1.$

Thus, if we introduce $\displaystyle x=\frac{[\Delta APB]}{[\Delta ABC]},\,$ $\displaystyle y=\frac{[\Delta BPC]}{[\Delta ABC]},\,$ $\displaystyle z=\frac{[\Delta CPA]}{[\Delta ABC]},\,$ then $x,y,z\gt 0\,$ and $x+y+z=1.$

Define function $f:\,(0,\infty)\rightarrow\mathbb{R}\,$ as $\displaystyle f(x)=\left(x+\frac{1}{x}\right)^2.\,$ Then

$\displaystyle f'(x)=2\left(x+\frac{1}{x}\right)\left(x-\frac{1}{x^2}\right)=2x-\frac{2}{x^3}.$

And $\displaystyle f''(x)=2+\frac{6}{x^4}\ge 0,\,$ making function $f$ convex and Jensen's inequality applicable:

\displaystyle\begin{align}f(x)+f(y)+f(z)&\ge \displaystyle 3f\left(\frac{x+y+z}{3}\right)\\ &=3f\left(\frac{1}{3}\right)\\ &=3\left(3+\frac{1}{3}\right)^2=\frac{100}{3}. \end{align}

Thus,

\displaystyle\begin{align}\sum_{P\in\{O,I,G\}}\sum_{cycl}\left(\frac{[\Delta APB]}{[\Delta ABC]}+\frac{[\Delta ABC]}{[\Delta APB]}\right)^2&\ge \sum_{P\in\{O,I,G\}}\frac{100}{3}\\ &=100. \end{align}

Remark

The restriction to acute triangles would have been superfluous, if it were not for one of the points involved being the circumcenter $O\,$ which lies in the exterior of obtuse triangles. This makes at least one of the variables $x,y,z\,$ negative. If the problem stipulates that all three selected points are located in the interior of the triangle, the inequality become true for every triangle. For example, assume $\Delta RST\,$ is the Morley triangle of $\Delta ABC.\,$ Then

$\displaystyle \sum_{P\in\{R,S,T\}}\sum_{cycl}\left(\frac{[\Delta APB]}{[\Delta ABC]}+\frac{[\Delta ABC]}{[\Delta APB]}\right)^2\ge 100.$

Acknowledgment

The problem (from the Romanian Mathematical Magazine) has been posted by Dan Sitaru at the CutTheKnotMath facebook page, Dan later communicated his solution on a LaTeX file by email.