China Western Mathematical Olympiad

(1) In the first diagram, if $P\in AC$, we have $PA\cdot PC\le \frac{1}{4}$ and $PB\le \sqrt{2}$. Thus $PA\cdot PB\cdot PC\le \frac{\sqrt{2}}{4}$. The equality is not valid, since the two equality signs cannot be valid at the same time. Therefore $PA\cdot PB\cdot PC<\frac{\sqrt{2}}{4}$.



(2) In the second diagram, if $P\in AB$, write $AP=x\in [0,\sqrt{2}]$, then



Let $t=x(\sqrt{2}-x)$, then $t\in [0,\frac{1}{2}]$ and $f(x)=g(t)=t^2(1-t)$.

Note that $g^{\prime }(t)=2t-3t^2=t(2-3t)$. Thus $g(t)$ is increasing on $[0,\frac{2}{3}]$ and $f(x)\le g(\frac{1}{2})=\frac{1}{8}$. Therefore $PA\cdot PB\cdot PC\le \frac{1}{2\sqrt{2}}=\frac{\sqrt{2}}{4}$. The equality is valid if and only if $t=\frac{1}{2}$ and $x=\frac{\sqrt{2}}{2}$. So $P$ is the midpoint of $AB$.