Question

Cho tam giác \(ABC.\) Chứng minh rằng:

\(\cos A+\cos B+\cos C+\dfrac{1}{\sin A}+\dfrac{1}{\sin B}\) \(+\dfrac{1}{\sin C}\ge2\sqrt{3}+\dfrac{3}{2}\)

Answer 1

Đặt \(f\left(A,B,C\right)=cosA+cosB+cosC+\dfrac{1}{sinA}+\dfrac{1}{sinB}+\dfrac{1}{sinC}-2\sqrt{3}-\dfrac{3}{2}\)

Ta có: \(f\left(A,B,C\right)-f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)\)

\(=\left(cosB+cosC-2cos\left(\dfrac{B+C}{2}\right)\right)+\left(\dfrac{1}{sinB}+\dfrac{1}{sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}\right)\)

\(=2cos\left(\dfrac{B+C}{2}\right)\left(cos\left(\dfrac{B-C}{2}\right)-1\right)+\left(\dfrac{1}{sinB}+\dfrac{1}{sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}\right)\left(1\right)\)

Bên cạnh đó ta có:

\(\dfrac{1}{sinB}+\dfrac{1}{sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}\ge\dfrac{4}{sinB+sinC}-\dfrac{2}{sin\left(\dfrac{B+C}{2}\right)}=\dfrac{4\left(1-cos\left(\dfrac{B-C}{2}\right)\right)}{sinB+sinC}\)

Do đó \(\left(1\right)\ge2\left(1-cos\left(\dfrac{B-C}{2}\right)\right)\left(\dfrac{2}{sinB+sinC}-cos\left(\dfrac{B+C}{2}\right)\right)\)

\(=\left(1-cos\left(\dfrac{B-C}{2}\right)\right)\left(\dfrac{1-sin\left(\dfrac{B+C}{2}\right)cos\left(\dfrac{B+C}{2}\right)cos\left(\dfrac{B-C}{2}\right)}{sinB+sinC}\right)\ge0\)

\(\Rightarrow f\left(A,B,C\right)\ge f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)\)

Giờ ta chỉ cần chứng minh bất đẳng thức đúng trong trường hợp tam giác cân.

Ta có: \(\left\{{}\begin{matrix}B=\dfrac{\pi}{2}-\dfrac{A}{2}\\cosB=cosC=\dfrac{sinA}{2}\\sinB=sinC=\dfrac{cosA}{2}\end{matrix}\right.\)

\(f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)=\left(cosA+2sin\left(\dfrac{A}{2}\right)-\dfrac{3}{2}\right)+\left(\dfrac{1}{sinA}+\dfrac{2}{cos\left(\dfrac{A}{2}\right)}-2\sqrt{3}\right)\)

\(=\dfrac{-2\left(sin\left(\dfrac{A}{2}\right)-1\right)^2}{2}+\dfrac{1+4sin\left(\dfrac{A}{2}\right)-2\sqrt{3}sinA}{sinA}\)

Mà ta có: \(1\ge sin\left(\dfrac{A}{2}+\dfrac{\pi}{3}\right)\)

\(\Rightarrow8sin\left(\dfrac{A}{2}\right)\ge2\sqrt{3}sinA+4sin^2\left(\dfrac{A}{2}\right)\)

\(\Rightarrow1+4sin\left(\dfrac{A}{2}\right)-2\sqrt{3}sinA\ge4sin^2\left(\dfrac{A}{2}\right)-4sin\left(\dfrac{A}{2}\right)+1=\left(2sin\left(\dfrac{A}{2}-1\right)\right)^2\)

Từ đó ta suy ra:

\(f\left(A,\dfrac{B+C}{2},\dfrac{B+C}{2}\right)\ge\left(2sin-1\right)^2\left(\dfrac{1}{sinA}-\dfrac{1}{2}\right)\ge0\)

Vậy bài toán đã được chứng minh. Dấu = xảy ra khi \(A=B=C=\dfrac{\pi}{3}\)

Answer 2

Hàm số \(f\left(x\right)=\cos\left(x\right)+\dfrac{1}{\sin\left(x\right)}\) là hàm lồi trên \(\left(0,\pi\right)\)

Do đó theo BĐT Jensen ( trường hợp của Karamata) có:

\(f\left(A\right)+f\left(B\right)+f\left(c\right)\ge3f\left(\dfrac{A+B+C}{3}\right)=3f\left(\dfrac{\pi}{3}\right)=2\sqrt{3}+\dfrac{3}{2}\)

P/s:Tính độ "lầy" của hàm số:

\(f''(x)=-\cos(x)-\frac{1}{\sin(x)}+\frac{2}{(\sin(x))^3}\)

Và cho \(x\in (0,\pi);f''(x)>0\) nếu \(2>(\sin(x))^2(\sin(x)\cos(x)+1)\) là xài dc Jensen :D