$M$ 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 $$S'\leq \frac{1}{3}S$$

 A: 
We can assume that side of a triangle $ABC$ is $1$. Further let $CM =x$ and $\angle KCM =\gamma$. Rotate $M$ around $C$ for $-60^{\circ}$ in to $F$. Then the area of triangle $AMF$ is the one we are looking for and it's area is area of $AMCF$ minus area of equilateral triangle $CFM$, so $$4S' = -x^2\sqrt{3}+2x\sin (60^{\circ}+\gamma)$$
and this should be easy to calculate that is less than ${\sqrt{3}\over 3}$.

If we see $S'$ quadratic function on $x$ we get:
$$ 4S'\leq {1\over \sqrt{3}}\sin (60^{\circ}+\gamma)\leq {1\over \sqrt{3}}$$
From here we can see that equality is achieved iff $\gamma = 30^{\circ}$ and $x= {\sqrt{3}\over 3} = {2\over 3}v$ where $v$ is altitude of triangle $ABC$. That is, equality is achieved iff $M$ is gravity centre of $ABC$.
A: 
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 $\sum MA^2 \ge 4\sqrt3 P$.
A: Choose a coordinate system so that triangle $ABC$ is lying on the unit circle centered at origin and $A$ on the $x$-axis. Let $a = AM$, $b = BM$, $c = CM$ and $S'$ be the area of a triangle with sides $a,b,c$. In this coordinate system, $S = \frac{3\sqrt{3}}{4}$, we want to show $S' \le \frac{\sqrt{3}}{4}$. Using Heron's formula, this is equivalent to
$$16S'^2 = (a^2+b^2+c^2)^2 - 2(a^4+b^4+c^4) \stackrel{?}{\le} 3$$
Identify euclidean plane with the complex plane. The vertices $A,B,C$ corresponds to $1, \omega, \omega^2 \in \mathbb{C}$ where $\omega = e^{\frac{2\pi}{3}i}$ is the cubic root of unity. Let $z$ be the complex number corresponds to $M$ and $\rho = |z|$, we have
$$
\begin{cases}
a^2 = |z-1|^2 = \rho^2 + 1  - (z + \bar{z})\\
b^2 = |z-\omega|^2 = \rho^2 + 1 - (z\omega + \bar{z}\omega^2)\\
c^2 = |z-\omega^2|^2 = \rho^2 + 1 - (z\omega^2 + \bar{z}\omega)
\end{cases}
\implies
a^2 + b^2 + c^2 = 3(\rho^2 + 1)
$$
Thanks to the identity $\omega^2 + \omega + 1 = 0$, all cross terms involving $\omega$ explicitly get canceled out.   
Doing the same thing to $a^4 + b^4 + c^4$, we get
$$\begin{align}a^4 + b^4 + c^4 
&= \sum_{k=0}^2 (\rho^2 + 1 + (z\omega^k + \bar{z}\omega^{-k}))^2\\
&= \sum_{k=0}^2\left[ (\rho^2 + 1)^2 + (z\omega^k + \bar{z}\omega^{-k})^2\right]\\
&= 3(\rho^2 + 1)^2 + 6\rho^2\end{align}$$
Combine these, we obtain
$$16S'^2 = 3(\rho^2+1)^2 - 12\rho^2 = 3(1 - \rho^2)^2$$
Since $M$ is inside triangle $ABC$, we have $\rho^2 \le 1$. As a result,
$$S' = \frac{\sqrt{3}}{4}(1-\rho^2) \le \frac{\sqrt{3}}{4} = \frac13 S$$
