<s>[INST] $M$ is a point in an equaliateral $ABC$ of area $S$. $S'$ is the area of the triangle with sides $MA,MB,MC$. Prove that $S'\leq \frac{1}{3}S$. 
$M$ is a point in an equilateral triangle $ABC$ with the area $S$. Prove that $MA, MB, MC$ are the lengths of three sides of a triangle which has area $$S'\leq \frac{1}{3}S$$

 [/INST] 
Let $N$ be the rotation of $M$ about $C$ by an angle of $60^\circ$, then $\triangle AMN$ has three sides equal to $MA,MB, MC$ respectively. Now
$$\begin{aligned}
P &= \frac12 MN\cdot MA\cdot \sin(\angle AMN)\\
&= \frac12 MC \cdot MA\cdot\sin(\angle AMC - 60)\\
&= \frac12 MC \cdot MA\cdot (\sin(\angle AMC)\cos 60 - \cos(\angle AMC)\sin 60)\\
&= \frac12 S_{MCA} - \frac{\sqrt3}{4}MC\cdot MA\cdot \cos(\angle AMC),
\end{aligned}$$
where $P$ denotes the area of $\triangle AMN$. The cosine law in $\triangle AMC$ gives us
$$MC\cdot MA \cdot \cos(\angle AMC) = \frac12 (MA^2+MC^2 - a^2),$$
where $a$ is the side of $\triangle ABC$. Substitute that to the above, we have
$$ P = \frac12 S_{MCA} - \frac{\sqrt3}8(MA^2 +MC^2 - a^2).$$
Take cyclic sum of the above with respect to $A,B,C$, we get
$$3P = \frac12S - \frac{\sqrt3}8\sum_{cyclic}(MA^2 + MC^2 -a^2),$$
where $S$ is the area of $\triangle ABC$, which is also equal to $\frac{\sqrt3}4a^2$. Thus we have
$$\begin{aligned}
3P &= \frac12S + \frac{3\sqrt3}8a^2 - \frac{\sqrt3}4\sum MA^2\\
&= 2S - \frac{\sqrt3}4\sum MA^2.
\end{aligned}$$
The fact $P\le \frac13S$ follows from the fact that $\sum MA^2 \ge 4\sqrt3 P$.
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