Prove the inequality $4S \sqrt{3}\le a^2+b^2+c^2$ Let a,b,c be the lengths of a triangle, S - the area of the triangle. Prove that
$$4S \sqrt{3}\le a^2+b^2+c^2$$ 
 A: Let $\triangle$ denote the area.
$\triangle=\sqrt{s(s-a)(s-b)(s-c)}$
Take $b+c-a=x$
$c+a-b=y$ and $a+b-c=z$.
$\dfrac{x}{2}=(s-a)$
$\dfrac{y}{2}=(s-b)$
$\dfrac{z}{2}=(s-c)$
$\dfrac{x+y+z}{2}=s$
Now use AM-GM inequality.
$\dfrac{x+y+z}{4} \ge ((xyz)(x+y+z))^\frac{1}{4}$
$\dfrac{2}{x+y+z} \le \dfrac{1}{((xyz)(x+y+z))^\frac{1}{4}}$
$\dfrac{4}{(x+y+z)^2} \le \dfrac{1}{((xyz)(x+y+z))^\frac{1}{2}} \implies \dfrac{4}{(x+y+z)^2} \le \dfrac{1}{\triangle}$
A: $\dfrac{S^{2}}{s}=(s-a)(s-b)(s-c)\leq \left(\dfrac{(s-a)+(s-b)+(s-c)}{3}\right)^{3}=\dfrac{s^{3}}{27}$
$\therefore$ $S\leq\dfrac{s^{2}}{3\sqrt{3}}=\dfrac{(a+b+c)^{2}}{12\sqrt{3}}\leq\dfrac{1}{12\sqrt{3}}\cdot 3(a^{2}+b^{2}+c^{2})$
A: You can use the Law of cosine and the area formula of a triangle to solve this problem. Suppose the three angles are $A,B,C$ opposite to the sides $a,b,c$, respectively. Then
$$ c^2=a^2+b^2-2ab\cos C, S=\frac{1}{2}ab\sin C $$
and hence
\begin{eqnarray*}
a^2+b^2+c^2-4S\sqrt{3}&=&2[a^2+b^2-ab(\cos C+\sqrt{3}\sin C)]\\
&=&2[a^2+b^2-2ab\sin(\frac{\pi}{6}+C)]\\
&\ge&2(a^2+b^2-2ab)\\
&=&2(a-b)^2\\
&\ge&0
\end{eqnarray*}
since $\sin(\frac{\pi}{6}+C)\le 1$. The equal sign holds iff $a=b$ and $\sin(\frac{\pi}{6}+C)=1$, which implies that $a=b, C=\frac{\pi}{3}$ or $a=b=c$.
A: Let $A=\bigl(-{c\over2},0\bigr)$, $B=\bigl({c\over2},0\bigr)$, and $C=(u,h)$ with $h>0$.  If $c$ and $Q:=a^2+b^2+c^2$ are given we want to make the area $S$  as large as possible by suitably choosing $a$ and $b$. Since $S={1\over2}c\> h$ this means that we should make $h$ as large as possible. Now
$$b^2=\bigl({c\over2}+u\bigr)^2 + h^2\ ,\qquad a^2= \bigl({c\over2}-u\bigr)^2 + h^2\ ,$$
and therefore
$$2h^2=a^2+b^2-\bigl({c\over2}+u\bigr)^2-\bigl({c\over2}-u\bigr)^2=Q-{3\over2}c^2-2u^2\ .$$
From this we conclude that, given $c$ and $Q$, the area $S$ is maximal iff $u=0$, which is equivalent to $a=b$.
We assume without proof that, given $Q$, there is a triangle with maximal area. By the foregoing  this triangle cannot have two unequal sides; so it has to be equilateral, and it will then have the area $$S={Q\over 4\sqrt{3}}\ .$$
A: This is Weitzenböck's inequality.
https://en.wikipedia.org/wiki/Weitzenb%C3%B6ck%27s_inequality
