看題主說原題是 a+b≥2c ,求證 C≤\frac{\pi}3=60° ?
好吧這樣就清楚多了
首先有正弦定理 \frac a {sinA}=\frac b{sinB}=\frac c{sinC}=2R
代入題中不等式: 2RsinA+2RsinB\geq 2RsinC\Rightarrow sinA+sinB\geq2sinC
左邊和差化積 2sin(\frac{A+B}{2})cos(\frac{A+B}{2}) ≥2sinC
然後 A+B=180°-C\Rightarrow cos(\frac{C}{2})cos(\frac{A+B}{2}) ≥sinC=2sin(\frac{C}{2})cos(\frac{C}{2})
於是 cos(\frac{A+B}{2}) ≥2sin(\frac{C}{2})
不妨設 A\geq B\Leftrightarrow a\geq b
則 0< cos(\frac{A+B}{2})\leq1\Rightarrow0<sin(\frac{C}{2})\leq\frac{1}{2}
因為 0°<C<180°
所以 0°<\frac{C}{2}≤30°\Rightarrow0°<C≤60° ,得證