Let P be any point inside a acute angled triangle, prove that PA.PB.AB+ PB. PC .BC +PC .PA .CA> AB .BC. CA. also equality holds if P is orthocenter.

Question

Let $P$ be any point inside an acute angled triangle, prove that $$PA \cdot PB \cdot AB + PB \cdot PC \cdot BC +PC \cdot PA \cdot CA \ge AB\cdot BC\cdot CA$$ and equality holds if $P$ is orthocenter. I tried with vectors taking $P$ as zero vector. But after that nothing better came out. Tried using cosines and sine, but again not much useful. Any hint?

Answer 1

Choose a coordinate system such that $P$ is the origin. WOLOG, we will assume $A,B,C$ are positioned counterclockwisely with respect to $P$.

Identity the Euclidean plane with the complex plane. Let $a, b, c$ be the complex numbers corresponds to $A,B,C$ respectively.

Notice

$$ab(b-a) + bc(c-b) + ca(a-c) = (b-a)(c-a)(c-b)\tag{*1}$$ By triangle inequality, we get

$$|a||b||a-b| + |b||c||c-b| + |c||a||a-c| \ge |b-a||c-a||c-b| \tag{*2}$$ In terms of the distances among the points, this reduces to the desired inequality $$PA\cdot PB \cdot AB + PB \cdot PC \cdot BC + PC \cdot PA \cdot CA \ge AB\cdot AC \cdot BC $$

When $P$ is orthocenter of $\triangle ABC$, we have $PA \perp BC$, $PB \perp CA$ and $PC \perp AB$. Since $A,B,C$ are positioned counterclockwisely with respect to $P$, following three numbers

$$i\frac{c-b}{a},\;\;i\frac{a-c}{b},\;\;i\frac{b-a}{c}$$

need to be positive real numbers. This means the three terms on LHS of $(*1)$ are positive multiples of each other. As a result,the inequality in $(*2)$ becomes an equality.

As a side note, the condition on equality is an if and only if one.

When $(*2)$ becomes an equality, the three terms on LHS of $(*1)$ need to be positive multiples of each other. We can find a complex number $\omega$ on the unit circle and three positive numbers $\lambda, \mu, \nu$ such that

$$\frac{c-b}{a} = \lambda\omega,\;\; \frac{a-c}{b} = \mu\omega,\;\; \frac{b-a}{c} = \nu\omega$$

Substitute these back into $(*1)$, we get $$abc\omega(\lambda + \mu + \nu) = -abc\lambda\mu\nu\omega^3 \implies \omega^2 = -\frac{\lambda+\mu+\nu}{\lambda\mu\nu} < 0$$ Since $A,B,C$ are positioned counterclockwisely with respect to $P$ and $|\omega| = 1$, we find $\omega = -i$ . From these, we can deduce $PA \perp BC$, $PB \perp CA$ and $PC \perp AB$. This means $P$ lies on the 3 altitudes of $\triangle ABC$ and is the orthocenter.

Up to my knowledge, this inequality is discovered by T. Hayashi in early 1910s.

• T. Hayashi, Two theorems on complex number, Tôhoku Math. J., 4 (1913-1914), 68–70.