What type of triangle satisfies: $8R^2 = a^2 + b^2 + c^2 $? In a $\displaystyle\bigtriangleup$ ABC,R is circumradius and $\displaystyle 8R^2 = a^2 + b^2 + c^2 $ , then $\displaystyle\bigtriangleup$ ABC is of which type ?

In general, in $\triangle{ABC}$
If $\sin^2 A + \sin^2 B + \sin^2 C \gt 2$ then $\triangle{ABC}$ is acute angled.
If $\sin^2 A + \sin^2 B + \sin^2 C  = 2$ then $\triangle{ABC}$ is right angled.
If $\sin^2 A + \sin^2 B + \sin^2 C \lt 2$ then $\triangle{ABC}$ is obtuse angled.
Assume $A \le B \lt \pi/2$ and $ A \le B \le C$.
Basically, if $k = \sin^2 A + \sin^2 B + \sin^2 C$
then we have that
$3-2k = \cos 2A + \cos 2B + \cos (2A+2B)$
i.e
$3-2k = 2\cos(A+B)\cos(A-B) + 2\cos^2(A+B) -1 $
i.e
$4-2k = 4\cos(A+B)\cos A\cos B$
So if $k > 2$, then $\cos(A+B) \lt 0$ hence acute.
$k = 2$, then $\cos(A+B) = 0$ hence right triangled.
$k < 2$, then $\cos(A+B) \gt 0$, hence obtuse.
In fact, we can go further and show that the maximum possible value of $k$ is $k = \frac{9}{4}$ which corresponds to $\triangle{ABC}$ being equilateral, as follows:
$4\sin^2 A + 4\sin^2 B + 4\sin^2 C = 9  + \delta$
i.e.
$(2 - 2\cos2A) + (2-2\cos 2B) + 4(1- \cos^2 (A+B)) = 9 + \delta$
i.e.
$1 + 2\cos2A + 2\cos 2B + 4\cos^2(A+B) = -\delta$
i.e.
$1 + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$
i.e.
$\sin^2(A-B) + \cos^2(A-B) + 4\cos(A+B)\cos(A-B) + 4\cos^2(A+B) = -\delta$
i.e.
$\sin^2(A-B) + (\cos (A-B) + 2\cos(A+B))^2 = -\delta$.
Hence $\delta \le 0$ and so $\sin^2 A + \sin^2 B + \sin^2 C \le \frac{9}{4}$
The case $\delta = 0$ gives us $\sin(A-B) = 0$ and $\cos(A+B) = \frac{-1}{2}$.
Hence $A=B=C$.
Thus the max value of $\sin^2 A + \sin^2 B + \sin^2 C$ is $\frac{9}{4}$ and is achieved when $A=B=C$.
