Gọi $x$, $2x$, $4x$ (rad) là số đo ba góc $A$, $B$, $C$ ($x>0$)
$\to x+2x+4x=\pi$
$\to x=\dfrac{\pi}{7}$
$\to A=\dfrac{\pi}{7}; B=\dfrac{2\pi}{7}; C=\dfrac{4\pi}{7}$
Ta có $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R$
$a^2+b^2+c^2$
$=\Big(2R\sin\dfrac{\pi}{7}\Big)^2+\Big(2R\sin\dfrac{2\pi}{7}\Big)^2+\Big(2R\sin\dfrac{4\pi}{7}\Big)^2$
$=4R^2.\Big( \sin^2\dfrac{\pi}{7}+\sin^2\dfrac{2\pi}{7}+\sin^2\dfrac{4\pi}{7}\Big)$
Do đó cần chứng minh đẳng thức: $\sin^2\dfrac{\pi}{7}+\sin^2\dfrac{2\pi}{7}+\sin^2\dfrac{4\pi}{7}=\dfrac{7}{4}$
Thật vậy:
$VT=\sin^2\dfrac{\pi}{7}+\sin^2\dfrac{2\pi}{7}+\sin^2\dfrac{4\pi}{7}$
$=\dfrac{1}{2}-\dfrac{1}{2}\cos\dfrac{2\pi}{7}+\dfrac{1}{2}-\dfrac{1}{2}\cos\dfrac{4\pi}{7}+\dfrac{1}{2}-\dfrac{1}{2}\cos\dfrac{8\pi}{7}$
$=\dfrac{3}{2}-\dfrac{1}{2}\Big( \cos\dfrac{2\pi}{7}+\cos\dfrac{4\pi}{7}+\cos\dfrac{8\pi}{7}\Big)$
$=\dfrac{3}{2}-\dfrac{1}{2}\Big(\dfrac{ 2\sin\dfrac{ \pi}{7}.\cos\dfrac{2\pi}{7}+2\cos\dfrac{4\pi}{7}\sin\dfrac{ \pi}{7}+2\sin\dfrac{ \pi}{7}\cos\dfrac{8\pi}{7} }{ 2\sin\dfrac{ \pi}{7}} \Big)$
$=\dfrac{3}{2}-\dfrac{1}{2}\Big( \dfrac{ \sin\dfrac{3\pi}{7}-\sin \dfrac{\pi}{7}+ \sin\dfrac{5\pi}{7}-\sin \dfrac{3\pi}{7}+\sin\dfrac{9\pi}{7} -\sin\pi }{2\sin\dfrac{\pi}{7}}\Big)$
$= \dfrac{3}{2}-\dfrac{1}{2}\Big( \dfrac{ -\sin\dfrac{\pi}{7}+\sin\dfrac{5\pi}{7}+ \sin\dfrac{9\pi}{7} }{2\sin\dfrac{\pi}{7}}\Big)$
$=\dfrac{3}{2}-\dfrac{1}{2}\Big( \dfrac{ -\sin\dfrac{\pi}{7}+\sin\dfrac{5\pi}{7}-\sin\dfrac{5\pi}{7} }{2\sin\dfrac{\pi}{7}}\Big)$
$=\dfrac{3}{2}-\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{7}{4}$
Vậy: $a^2+b^2+c^2=4R^2.\dfrac{7}{4}=7R^2$ (đpcm)
