<s>[INST] For a triangle $ABC$, $a^2+b^2+c^2=8R^2$ then it is a right triangle? $ABC$ is a triangle, $a^2+b^2+c^2=8R^2$ then how do we prove it is a right triangle?
 [/INST] Let $S$ be the area of the triangle.
Since we have $4RS=abc$ and
$$\begin{align}16S^2&=(a+b+c)(-a+b+c)(a-b+c)(a+b-c)\\&=-(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+c^2a^2)\end{align}$$
if $a^2+b^2+c^2=8R^2$, then we have
$$\begin{align} 0 &=\frac{a^2+b^2+c^2}{R^2}-8\\&=\frac{a^2+b^2+c^2}{a^2b^2c^2}(-a^4-b^4-c^4+2a^2b^2+2b^2c^2+2c^2a^2)-8\\&=\frac{(-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)}{a^2b^2c^2}.\end{align}$$
$$\Rightarrow (-a^2+b^2+c^2)(a^2-b^2+c^2)(a^2+b^2-c^2)=0$$
So, the triangle has to be a right triangle.
P.S. Actually, $a^2+b^2+c^2=8R^2\iff\text{The triangle $ABC$ is a right triangle}$.
Also, we can see the followings :
$$a^2+b^2+c^2\gt 8R^2\iff \text{The triangle $ABC$ is an acute‐angled triangle}.$$
$$a^2+b^2+c^2\lt 8R^2\iff \text{The triangle $ABC$ is an obtuse triangle}.$$
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