$1)\sin A+\sin B+\sin C\leq \frac{3\sqrt{3}}{2}$
$2)\cos A+\cos B+\cos C\leq \frac{3}{2}$          ABC là tam giác nhọn
$3)\tan A+\tan B+\tan C\geq 3\sqrt{3}$           ABC là tam giác nhọn
$4)\cot A+\cot B+\cot C\geq \sqrt{3}$              ABC là tam giác nhọn
$5)\sin^2A+\sin^2B+\sin^2C\leq \frac{9}{4}$
$6)\sin A\sin B\sin C\leq \frac{3\sqrt{3}}{8}$
$7)\cos A\cos B\cos C\leq \frac{1}{8}$
$8)\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\leq \frac{1}{8}$
$9)\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}\leq \frac{3\sqrt{3}}{8} $
$10)\tan\frac{A}{2}\tan\frac{B}{2}\tan\frac{C}{2}\leq \frac{1}{3\sqrt{3}}$

10 Đáp án

Câu 6)
Lấy (8) nhân (9)=> (6)
Lấy $\frac{(8)}{(9)}\Rightarrow (10)$
Câu 9)
$\sin A+\sin B+\sin C=4VT(9)\leq \frac{3\sqrt{3}}{2}\Rightarrow $ đpcm
Cách làm tương tự câu 7 ta cm được câu 8
Ta có $\sin^2A+\sin^2B+\sin^2C=2+2\cos A\cos B\cos C$
Vậy từ câu $7\Rightarrow 5$ 
Đặt $k=\cos A\cos B\cos C$
       $\Leftrightarrow 2k=(\cos (A-B)-\cos C)\cos C$
       $\Leftrightarrow \cos^2C-2\cos C\cos(A-B)+2k=0      (*)$
$(*)$ có nghiệm nên $\Delta \geq 0\Leftrightarrow \cos^2(A-B)\geq 8k\Leftrightarrow 8k\leq \cos^2(A-B)\leq 1\Rightarrow $ đpcm
Xét $f(x)=\cot x,     x\in (0;\frac{\pi}{2})$
$+f'(x)=-\frac{1}{\sin^2x}$
$+f''(x)=\frac{\sin 2x}{\sin^4x}>0\Rightarrow f(x)$ lõm trên $(0;\frac{\pi}{2})$
Theo Jensen => đpcm
Xét $f(x)=\tan x,      x\in (0;\frac{\pi}{2})$
$+f'(x)=\frac{1}{\cos^2x}$
$+f''(x)=\frac{\sin 2x}{cos^4x}>0\Rightarrow f(x)$ lõm trên $(0;\frac{\pi}{2})$
Theo Jensen ta có được đpcm
Tương tự: Xét $f(x)=\cos x,   0<x<\frac{\pi}{2}$
$+f'(x)=-\sin x$
$+f''(x)=-\cos x<0\Rightarrow f(x)$ lồi trên $(0;\frac{\pi}{2})$
Cũng theo Jensen: $f(A)+f(B)+f(C)\leq 3f(\frac{A+B+C}{2})$
                             $\Leftrightarrow \cos A+\cos B+\cos C\leq 3\cos\frac{\pi}{3}=\frac{3}{2}$  (đpcm)

1)Xét $f(x)=\sin x,      0<x<\pi$
$+f'(x)=\cos x$
$+f''(x)=-\sin x<0\Rightarrow f(x)$ lồi trên $(0;\pi)$
Theo Jensen: $f(A)+f(B)+f(C)\leq 3f(\frac{A+B+C}{3})$
                    $\Leftrightarrow \sin A+\sin B+\sin C\leq 3\sin\frac{\pi}{3}=\frac{3\sqrt{3}}{2}$ đpcm

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