Đáp án:
Giải thích các bước giải:
a) Không mất tính tổng quát giả sử: \(A\ge B\ge C\). Khi đó \(A\ge\dfrac{\pi}{3};C\le\dfrac{\pi}{3}\)
Vì \(\dfrac{\pi}{2}\ge A\ge\dfrac{\pi}{3}\) và \(\pi\ge A+B=\pi-C\ge\dfrac{2\pi}{3}\) nên
\(\left\{{}\begin{matrix}\dfrac{\pi}{2}\ge A\ge\dfrac{\pi}{3}\\\dfrac{\pi}{2}+\dfrac{\pi}{2}\ge A+B\ge\dfrac{\pi}{3}+\dfrac{\pi}{3}\\\dfrac{\pi}{2}+\dfrac{\pi}{2}+0=A+B+C=\dfrac{\pi}{3}+\dfrac{\pi}{3}+\dfrac{\pi}{3}\end{matrix}\right.\)
Xét hàm số \(f\left(x\right)=\cos x\forall x\in\left[0;\dfrac{\pi}{2}\right]\)
Ta có: \(f"\left(x\right)=-\cos x< 0\forall x\in\left[0;\dfrac{\pi}{2}\right]\) nên hàm số \(f\left(x\right)\) lõm trên đoạn \(\left[0;\dfrac{\pi}{2}\right]\). Khi đó, theo BĐT Karamata ta có:
\(f\left(\dfrac{\pi}{2}\right)+f\left(\dfrac{\pi}{2}\right)+f\left(0\right)\le f\left(A\right)+f\left(B\right)+f\left(C\right)\le3f\left(\dfrac{\pi}{3}\right)\)
Hay \(\cos A+\cos B+\cos C\le\dfrac{3}{2}\)
b)
Ta có:
\(\begin{array}{l}
\cos 2x = 1 - 2{\sin ^2}x \Rightarrow {\sin ^2}x = \frac{{1 - \cos 2x}}{2}\\
A = {\sin ^2}A + {\sin ^2}B + {\sin ^2}C\\
= \frac{{1 - \cos 2A}}{2} + \frac{{1 - \cos 2B}}{2} + {\sin ^2}C\\
= 1 - \frac{{\cos 2A + \cos 2B}}{2} + \left( {1 - {{\cos }^2}C} \right)\\
= 1 - \frac{{2.cos\frac{{2A + 2B}}{2}.\cos \frac{{2A - 2B}}{2}}}{2} + 1 - {\cos ^2}C\\
= 2 - {\cos ^2}C - \cos \left( {A + B} \right).\cos \left( {A - B} \right)\\
= 2 - {\cos ^2}C - \left( { - \cos \left( {180^\circ - \left( {A + B} \right)} \right)} \right).\cos \left( {A - B} \right)\\
= 2 - {\cos ^2}C + \cos C.\cos \left( {A - B} \right)\\
- 1 \le \cos \left( {A - B} \right) \le 1\\
TH1:\,\,\,\,\left\{ \begin{array}{l}
- 1 \le \cos C \le 0\\
- 1 \le \cos \left( {A - B} \right) \le 1
\end{array} \right. \Rightarrow \cos C.\cos \left( {A - B} \right) \le - \cos C\\
\Rightarrow A \le 2 - {\cos ^2}C - \cos C = \frac{9}{4} - \left( {{{\cos }^2}C + \cos C + \frac{1}{4}} \right) = \frac{9}{4} - {\left( {\cos C + \frac{1}{2}} \right)^2} \le \frac{9}{4}\\
TH2:\,\,\,\,\,\left\{ \begin{array}{l}
0 \le \cos C \le 1\\
- 1 \le \cos \left( {A - B} \right) \le 1
\end{array} \right. \Rightarrow \cos C.\cos \left( {A - B} \right) \le \cos C\\
\Rightarrow A \le 2 - {\cos ^2}C + \cos C = \frac{9}{4} - \left( {{{\cos }^2}C - \cos C + \frac{1}{4}} \right) = \frac{9}{4} - {\left( {\cos C - \frac{1}{2}} \right)^2} \le \frac{9}{4}
\end{array}\)
c) `:. (sin^2(A/2)sin^2(B/2)sin^2(C/2))^(1/3) le 1/3(sin^2(A/2)+sin^2(B/2)+sin^2(C/2))->(1)`
Ta có, `sin^2(A/2)+sin^2(B/2)+sin^2(C/2) ge 3/4`
`:. (sin^2(A/2)sin^2(B/2)sin^2(C/2))^(1/3) le 1/3(3/4)`
`=>(sin^2(A/2)sin^2(B/2)sin^2(C/2))^(1/3) le 1/4`
`=>(sin^2(A/2)sin^2(B/2)sin^2(C/2)) le (1/4)^3`
`=>sin(A/2)sin(B/2)sin(C/2) le (1/64)^(1/2)`
`=>sin(A/2)sin(B/2)sin(C/2) le 1/8`
