$\sin A+\sin B+\sqrt{6}\sin C \leq \frac{5\sqrt{10}}{4}.$

Question

Chứng minh rằng với mọi

$\Delta ABC$ ta đều có:

score 11 · Answer 1

Ta có:

$P=\sin A+\sin B+\sqrt6\sin C$

$=2\sin\dfrac{A+B}{2}\cos\dfrac{A-B}{2}+\sqrt6\sin C$

$=2\cos\dfrac{C}{2}\cos\dfrac{A-B}{2}+\sqrt6\sin C$

$\le2\cos\dfrac{C}{2}+\sqrt6\sin C$

Khảo sát hàm số:

$f(x)=2\cos\dfrac{x}{2}+\sqrt6\sin x,x\in(0;\pi)$

$f'(x)=-\sin\dfrac{x}{2}+\sqrt6\cos x=-2\sqrt6\sin^2\dfrac{x}{2}-\sin\dfrac{x}{2}+\sqrt6$

$f'(x)=0 \Leftrightarrow \sin\dfrac{x}{2}=\dfrac{\sqrt6}{4}=\sin\dfrac{\alpha}{2}$ , với

$\alpha=2\arcsin\dfrac{\sqrt6}{4}$ .

Lập bảng biến thiên, suy ra:

$f(x)\le f(\alpha)=\dfrac{5\sqrt{10}}{4}$

Vậy

$P\le\dfrac{5\sqrt{10}}{4}$ .

Dấu bằng xảy ra khi và chỉ khi:

$\left\{\begin{array}{l}A=B\\C=2\arcsin\dfrac{\sqrt6}{4}\end{array}\right.$

1 Đáp án