Ta có: $sin^{2}A$ + $sin^{2}B$ + $sin^{2}C$ = 2 + 2.cosA.cosB.cosCMặt khác: Đặt k = coaA.cosB.cosC = $\frac{1}{2}$[cos(A-B) + cos(A+B)].cosC
$\Rightarrow $2k = cos(A-B)cosC - $cos^{2}C$ $\Leftrightarrow$ $cos^{2}C$ -cos(A-B).cosC -2k =0
vì pt này có nghiệm nên:
$\triangle $= $cos^{2}(A-B)$ - 8k $\geq $0
$\Rightarrow$ 8k $\leq$ $cos^{2}(A-B)$ $ \leq$ 1 $\Rightarrow$ k$\leq$ $\frac{1}{8}$ hay coaA.cosB.cosC $\leq $ $\frac{1}{8}$
$\Rightarrow$ $sin^{2}A$ + $sin^{2}B$ + $sin^{2}C$ $\leq$ $\frac{9}{4}$