Bài 105207

Question

Cho

$\triangle ABC$ chứng minh rằng:

$1< \sin{\frac{A}{2}}+ \sin{\frac{B}{2}}+ \sin{\frac{C}{2}} \leq \frac{3}{2}$

score 0 · Answer 1

a.Chứng minh vế trái, ta có:

$\sin{\frac{A}{2}}+ \sin{\frac{B}{2}}+ \sin{\frac{C}{2}}\geq \sin{\frac{A}{2}} \cos {\frac{B}{2}}+ \sin{\frac{B}{2}} \cos{\frac{A}{2}}+ \sin{\frac{C}{2}}$

$=\sin{\frac{A+B}{2}}+ \sin{\frac{C}{2}}= \cos {\frac{C}{2}} +\sin{\frac{C}{2}}$

$=\sqrt{2}\sin({{\frac{C}{2}}+{\frac{\pi}{4}}} )>\sqrt{2} \sin{\frac{\pi}{4}}=1$ .
b. Chứng minh vế phải
xét hàm số

$f(x)=\sin x$ với

$x\in (0,\frac{\pi}{2})$ , ta có:

$f^'(x)=\cos x, f^{''}(x)=-\sin x\leq0, \forall x\in (0,\frac{\pi}{2})$
Vậy hàm số

$f(x)=\sin x$ là lồi trên

$(0,\frac{\pi}{2})$ , do đó: với

$\forall \frac{A}{2},\frac{B}{2},\frac{C}{2}\in(0,\frac{\pi}{2})$ , ta có:

$\frac {\displaystyle f(\frac{A}{2})+f(\frac{B}{2})+f(\frac{C}{2})}{3}\leq f\frac{A+B+C}{6}\Leftrightarrow \frac{\displaystyle \sin{\frac{A}{2}}+ \sin{\frac{B}{2}}+ \sin{\frac{C}{2}}}{3}\leq \sin{\frac{\pi}{6}}$