I'm collecting proofs for Weitzenbock's inequality, once featured as a question in IMO 1961. I made three proofs for this. See these proofs below. (Whoever has a more cool proof, please share.)
Weitzenbock's inequality. Let $a$, $b$, $c$ be the sides of the triangle, and let $A$ be its area. Then:
$$a^{2}+b^{2}+c^{2}\;\geq\;4\sqrt{3}\;A$$
Proof 1
Let $R$ the circunradius. Suppose, by contradiction, that:
\begin{equation}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}<\frac{3}{R\sqrt{3}} \tag{1}
\end{equation}
Using that:
\begin{equation}
a+b+c\leq 3R\sqrt{3} \tag{2}
\end{equation}
Multiplying (1) e (2):
$$(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)<9$$
What is absurd, this can be seen using the inequality that relates the arithmetic mean to the harmonic mean, or by the inequality of Cauchy-Schwarz. From where we conclude:
\begin{equation}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq\frac{3}{R\sqrt{3}}
\end{equation}
Using $\displaystyle A=\frac{abc}{4R}$, we get:
$$ab+ac+bc\geq \frac{\sqrt{3}abc}{R}$$
$$ab+ac+bc\geq 4A\sqrt{3}$$
This result follows using that $\displaystyle a^{2}+b^{2}+c^{2}\geq ab+bc+ac$
Proof 2
By Heron we get:
\begin{equation*}
a^2+b^2+c^2 \geq 4\sqrt{3} \sqrt{S(S-a)(S-b)(S-c)}
\end{equation*}
Multiplying both sides of the inequality above by $\displaystyle \sqrt{\left(\frac{a+b+c}{2abc}\right)^2}=\sqrt{\left(\frac{S}{abc}\right)^2}$, :
\begin{equation*}
\left(a^2+b^2+c^2\right)\left(\frac{a+b+c}{2abc}\right) \geq 4\sqrt{3} \sqrt{\frac{S(S-a)}{bc}\frac{S(S-b)}{ac}\frac{S(S-c)}{ab}}
\end{equation*}
By the law of cosine and law of sine we have:
\begin{equation*}
\left(a^2+b^2+c^2\right)\left(\frac{a+b+c}{2abc}\right) \geq 4\sqrt{3} \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2}
\end{equation*}
\begin{equation*}
\left(\frac{a^2}{bc}+\frac{a}{b}+\frac{a}{c}\right)+ \left(\frac{b^2}{ac}+\frac{b}{a}+\frac{b}{c}\right)+ \left(\frac{c^2}{ab}+\frac{c}{a}+\frac{c}{b}\right) \geq 8\sqrt{3} \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2}
\end{equation*}
\begin{equation*}
\left(\frac{sen^2\alpha}{sen\beta sen\gamma}+\frac{sen\alpha}{sen\beta}+\frac{sen\alpha}{sen\gamma}\right)+\left(\frac{sen^2\beta}{sen\alpha sen\gamma}+\frac{sen\beta}{sen\alpha}+\frac{sen\beta}{sen\gamma}\right)+\left(\frac{sen^2\gamma}{sen\alpha sen\beta}+\frac{sen\gamma}{sen\alpha}+\frac{sen\gamma}{sen\beta}\right) \geq 8\sqrt{3} \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2}
\end{equation*}
\begin{equation*}
(sen\alpha+sen\beta+sen\gamma)\left(\frac{sen\alpha}{sen\beta sen\gamma}+\frac{sen\beta}{sen\alpha sen\gamma}+\frac{sen\gamma}{sen\alpha sen\beta}\right) \geq 8\sqrt{3} \cos\frac{\alpha}{2}\cos\frac{\beta}{2}\cos\frac{\gamma}{2}
\end{equation*}
\begin{equation*}
\cot\alpha+\cot\beta+\cot\gamma\geq \sqrt{3}
\end{equation*}
And this inequality follows from the Jensen inequality.
Proof 3
Let $\displaystyle R$ the circunradius, $\displaystyle \alpha, \beta, \gamma$ the opposite acute angles to the sides of $\displaystyle a, b, c$, so:
$\\ \\ \displaystyle a^2+b^2-c^2=\frac{4Rabc(a^2+b^2-c^2)}{4Rabc}=\frac{4AR(a^2+b^2-c^2)}{abc}=4A\times\frac{2R}{c}\times\frac{a^2+b^2-c^2}{2ab}=4Acot\gamma=4A\sqrt{cot^2\gamma}=4A\sqrt{csc^2\gamma-1}=4A\sqrt{\frac{4R^2}{c^2}-1}=4A\sqrt{\frac{4a^2b^2R^2}{a^2b^2c^2}-1}=4A\sqrt{\frac{a^2b^2}{4\left(\frac{abc}{4R}\right)^2}-1}=4A\sqrt{\frac{a^2b^2}{4A^2}-1}=2\sqrt{4A^2\left(\frac{a^2b^2}{4A^2}-1\right)}=2\sqrt{a^2b^2-4A^2}\Rightarrow a^2+b^2-c^2=2\sqrt{a^2b^2-4A^2}\\ \\$
By symmetry, we get:
\begin{equation*}
2\sqrt{a^2b^2-4A^2}=a^2+b^2-c^2
\end{equation*}
\begin{equation*}
2\sqrt{a^2c^2-4A^2}=a^2+c^2-b^2
\end{equation*}
\begin{equation*}
2\sqrt{b^2c^2-4A^2}=b^2+c^2-a^2
\end{equation*}
Multiplying these equalities two by two we will have:
\begin{equation*}
4\sqrt{a^2b^2-4A^2}\sqrt{b^2c^2-4A^2}=(a^2+b^2-c^2)(b^2+c^2-a^2)
\end{equation*}
\begin{equation*}
4\sqrt{a^2c^2-4A^2}\sqrt{b^2c^2-4A^2}=(a^2+c^2-b^2)(b^2+c^2-a^2)
\end{equation*}
\begin{equation*}
4\sqrt{a^2b^2-4A^2}\sqrt{a^2c^2-4A^2}=(a^2+c^2-b^2)(a^2+b^2-c^2)
\end{equation*}
We know that $\displaystyle m+n\geq 2\sqrt{mn}$, it folows that:
\begin{equation*}
2(a^2b^2-4A^2+b^2c^2-4A^2)\geq (a^2+b^2-c^2)(b^2+c^2-a^2)
\end{equation*}
\begin{equation*}
2(a^2c^2-4A^2+b^2c^2-4A^2)\geq(a^2+c^2-b^2)(b^2+c^2-a^2)
\end{equation*}
\begin{equation*}
2(a^2b^2-4A^2+a^2c^2-4A^2)\geq(a^2+c^2-b^2)(a^2+b^2-c^2)
\end{equation*}
Adding all the above inequalities we will have:
\begin{equation*}
4a^2b^2+4b^2c^2+4a^2c^2-16\times 3 A^2\geq (a^2+b^2-c^2)(b^2+c^2-a^2)+(a^2+c^2-b^2)(b^2+c^2-a^2)+(a^2+c^2-b^2)(b^2+c^2-a^2)
\end{equation*}
Let $ b^2+c^2-a^2=x,y=a^2+c^2-b^2,z=a^2+b^2-c^2$.Observe that
$$ 2a^2=(a^2+c^2-b^2)+(a^2+b^2-c^2)=y+z$$
$$ 2b^2=(a^2+b^2-c^2)+(b^2+c^2-a^2)=x+z$$
$$ 2c^2=(a^2+c^2-b^2)+(b^2+c^2-a^2)=x+y$$
We get:
$$ 4a^2b^2=(y+z)(x+z)$$
$$ 4a^2c^2=(y+z)(x+y)$$
$$ 4b^2c^2=(x+z)(x+y)$$
Replacing in the inequality above, we have:
\begin{equation*}
(x+z)(x+y)+(y+z)(x+y)+(y+z)(x+z)-16\times 3 A^2\geq xy+xz+yz
\end{equation*}
\begin{equation*}
x^2+y^2+z^2+2xy+2xz+2yz-16\times 3 A^2\geq 0
\end{equation*}
\begin{equation*}
(x+y+z)^2-16\times 3 A^2\geq 0
\end{equation*}
\begin{equation*}
x+y+z\geq 4\sqrt{3}A
\end{equation*}
Done
Best Answer
What about a proof (almost) without words?
Let us consider a triangle $ABC$ and its Fermat-Torricelli-Steiner point, such that $\widehat{CFA}=\widehat{AFC}=\widehat{CFB}=120^\circ$. It is well-known that $F$ lies on each line joining a vertex of $ABC$ with the free vertex of the equilateral triangle built on the opposite side. By replicating the (colored) triangles $AFC,CFB,BFA$ four times we reach the configuration depicted above. Due to the triangular holes $$ [ABV_C]+[ACV_B]+[BCV_A]\geq 3[ABC] $$ that is equivalent to the claim.
There also is an efficient variational approach. If $ABC$ is a triangle in the plane with $AB=c,AC=b$, the locus of points $P$ such that $PB^2+PC^2=b^2+c^2$ is a circle through $A$ centered at the midpoint of $BC$. It follows that by moving the vertex $A$ along such circle, till meeting the perpendicular bisector of $BC$, the area of $ABC$ increases but $AB^2+BC^2+AC^2$ stays the same. It follows that for a fixed $a^2+b^2+c^2$, the maximum area is achieved only by the equilateral triangle, and $a^2+b^2+c^2\geq 4\sqrt{3}\Delta$ follows.
Yet another way is to exploit the Cauchy-Schwarz inequality. We have $$ (1+1+1)(a^4+b^4+c^4)\geq (a^2+b^2+c^2)^2 $$ hence: $$ 4\Delta = \sqrt{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)} \leq \frac{1}{\sqrt{3}}(a^2+b^2+c^2). $$
Update. Yet another way is to prove a stronger inequality, namely $$ ab+ac+bc \geq 4\Delta\sqrt{3} $$ which follows from $ab=\frac{2\Delta}{\sin C}$ and the convexity of $\frac{1}{\sin\theta}$ over $(0,\pi)$.