[tex]\sum_{cyc} sinA=sinA+2.sin.\frac{B+C}{2}cos.\frac{B-C}{2}[/tex]
[tex]=sinA+2cos.\frac{A}{2}cos.\frac{B-C}{2} \leq 1+2.1.\frac{\sqrt{2}}{2}=1+2.\sqrt{2}[/tex]
[tex](gt) \Rightarrow b^2+c^2 \leq a^2 \Rightarrow \frac{\pi}{2} \leq A \leq \pi \Rightarrow \frac{\pi}{4} \leq \frac{A}{2} \leq \frac{\pi}{2}[/tex]
[tex] \Rightarrow \left{\begin{cos.\frac{A}{2} \leq \frac{\sqrt{2}}{2}}\\{cos.\frac{B-C}{2} \leq 1}[/tex]
[tex](gt) \Rightarrow \sum_{cyc} sinA =1+\sqrt{2}[/tex]
[tex]\rightarrow (ycbt) \leftrightarrow[/tex] đẳng thức xảy ra
[tex]\Rightarrow \left{\begin{cos.\frac{B-C}{2}=1}\\{sinA=1}\\{cos.\frac{A}{2}=\frac{\sqrt{2}}{2}[/tex]
[tex]\Rightarrow \left{\begin{A=\frac{\pi}{2}}\\{B=C=\frac{\pi}{4} [/tex]
Vậy ABC là tam giác vuông cân.
