trc hết cm: sina + sinb ≤ 2sin(a + b)/2
VT = 2sin(a+ b)/2.cos(a - b)/2
Do a ; b ; c Є [0 ; π] -> sin(a + b)/2 ≥ 0 ; cos(a - b)/2 ≤ 1
-> 2sin(a + b)/2.cos(a - b)/2 ≤ 2sin(a + b)/2 ( VP)
Dấu = xảy ra <=> cos(a - b)/2 = 1 <=> a = b
sinA+sinB+sinC3sin((A+B+C)/3)
áp dụng sina + sinb ≤ 2sin(a + b)/2
sinC + sin(a + b + c)/3 ≤ 2.sin[ c + (a + b + c)/3 ]/2
-> sina + sinb + sinc + sin(a + b + c)/3 ≤ 2sin(a + b)/2 + 2.sin[c + (a + b + c)/3]/2
≤ 2.[ sin(a + b)/2 + 2.sin[c + (a + b + c)/3]/2 ]
≤ 2.2.sin[ (a + b)/2 + [c + (a + b + c)/3]/2 ]/2 = 4sin(a + b + c)/3
-> sina + sinb + sinc + sin(a + b + c)/3 ≤ 4sin(a + b+ c)/3
-> sina + sinb + sinc ≤ 3sin(a + b + c)/3
- Dấu = xảy ra <=> a = b = c