[TEX]sin^{2}B+sin^{2}C=2sin^{2}A\Leftrightarrow (\frac{b}{2R})^{2}+(\frac{c}{2R})^{2}=2(\frac{a}{2R})^{2}[/TEX] (Định lí sin)
[TEX]\Leftrightarrow \frac{b^{2}+c^{2}}{4R^{2}}=\frac{2a^{2}}{4R^{2}} \Leftrightarrow b^{2}+c^{2}=2a^{2} [/TEX] (*)
[TEX]\Leftrightarrow \frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{a^{2}}{2bc}\Leftrightarrow cosBAC=\frac{a^{2}}{2bc}\geq \frac{a^{2}}{b^{2}+c^{2}}\, (AM-GM)[/TEX] (*)(*)
Theo (*) ta có: [TEX]b^{2}+c^{2}=2a^{2}\Leftrightarrow a^{2}=\frac{b^{2}+c^{2}}{2}[/TEX] (1)
Thay (1) vào (*)(*) ta được: [TEX]cosBAC\geq \frac{b^{2}+c^{2}}{2}.\frac{1}{b^{2}+c^{2}}=\frac{1}{2}\Rightarrow \hat{BAC}\leq 60^{0}[/TEX] (ĐPCM)