Ta có: sin^{2}A + sin^{2}B + sin^{2}C = 2 + 2xcosAxcosBxcosCMặt khác: Đặt k = coaAxcosBxcosC = \frac{1}{2} [cos(A-B) + cos(A+B)]xcosC\Leftrightarrow 2k = cos(A-B)cosC - cos^{2}C \Leftrightarrow cos^{2}C -cos(A-B)xcosC -2k =0vì 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 coaAxcosBxcosC\leq \frac{1}{8}\Rightarrow sin^{2}A + sin^{2}B + sin^{2}C \leq \frac{9}{4}

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 =0vì 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$ $sin^{2}A$ + $sin^{2}B$ + $sin^{2}C$ $\leq$ $\frac{9}{4}$