CM tam giác ABC đều nếu sina+sinb+sinc= cosa/2 + cosb/2 +cosc/2

Đáp án:

Giải thích các bước giải: Theo giả thiết:

$ cos\frac{A}{2} + cos\frac{B}{2} + cos\frac{C}{2} = sinA + sinB + sinC = 2sin\frac{A + B}{2}cos\frac{A – B}{2} + 2sin\frac{C}{2}cos\frac{C}{2} $

$= 2cos\frac{C}{2}cos\frac{A – B}{2} + 2cos\frac{C}{2}cos\frac{A + B}{2} = 2cos\frac{C}{2}(cos\frac{A – B}{2} + cos\frac{A + B}{2})$

$= 4cos\frac{A}{2}cos\frac{B}{2}cos\frac{C}{2} ≤ \frac{4}{27}(cos\frac{A}{2} + cos\frac{B}{2} + cos\frac{C}{2})³$ (Cô si cho 3 số)

$⇒ 1 ≤ \frac{4}{27}(cos\frac{A}{2} + cos\frac{B}{2} + cos\frac{C}{2})² ⇔ \frac{3\sqrt[]{3}}{2} ≤ cos\frac{A}{2} + cos\frac{B}{2} + cos\frac{C}{2} (1)$

Mặt khác áp dụng BĐT $: (x + y + z)² ≤ 3(x² + y² + z²)$ ta có :

$(cos\frac{A}{2} + cos\frac{B}{2} + cos\frac{C}{2})² ≤ 3(cos²\frac{A}{2} + cos²\frac{B}{2} + cos²\frac{C}{2})$

$= \frac{3}{2}(cosA + 1 + cosB + 1 + cosC + 1) = \frac{3}{2}(4 + cosA + cosB + cosC – 1)$

$= \frac{3}{2}(4 + 2cos\frac{A + B}{2}cos\frac{A – B}{2} – 2sin²\frac{C}{2}) = \frac{3}{4}(8 + 4cos\frac{A + B}{2}cos\frac{A – B}{2} – 4sin²\frac{C}{2}) $

$= \frac{3}{4}(9 – sin²\frac{A – B}{2} – cos²\frac{A – B}{2} + 4sin\frac{C}{2}cos\frac{A – B}{2} – 4sin²\frac{C}{2})$

$= \frac{3}{4}[9 – sin²\frac{A – B}{2} – (cos\frac{A – B}{2} – 2sin\frac{C}{2})²] ≤ \frac{27}{4} ⇔ cos\frac{A}{2} + cos\frac{B}{2} + cos\frac{C}{2} ≤ \frac{3\sqrt[]{3}}{2} (2)$

Từ $(1); (2) ⇒$ Đã xảy ra dấu $”=” ⇔  cos\frac{A}{2} = cos\frac{B}{2} = cos\frac{C}{2} = \frac{\sqrt[]{3}}{2} ⇔ A = B = C ⇔ ΔABC$ đều