§1. Bất đẳng thức

Đặt \(x=\dfrac{1}{a},y=\dfrac{1}{b},z=\dfrac{1}{c}\) khi đó thu được \(xyz=1\)

Ta có:

\(\dfrac{1}{a^2\left(b+c\right)}=\dfrac{x^2}{\dfrac{1}{y}+\dfrac{1}{z}}=\dfrac{x^2yz}{y+z}=\dfrac{x}{y+z}\)

BĐT cần chứng minh được viết lại thành:\(\dfrac{x}{y+z}+\dfrac{y}{z+x}+\dfrac{z}{x+y}\ge\dfrac{3}{2}\)

\(\Leftrightarrow\left(\dfrac{x}{y+z}+1\right)+\left(\dfrac{y}{z+x}+1\right)+\left(\dfrac{z}{x+y}+1\right)\ge\dfrac{9}{2}\)

\(\Leftrightarrow\left(x+y+z\right)\left(\dfrac{1}{y+z}+\dfrac{1}{z+x}+\dfrac{1}{x+y}\right)\ge\dfrac{9}{2}\)

Đánh giá cuối cùng đúng theo BĐT Cauchy

Vậy BĐT được chứng minh. Đẳng thức xảy ra khi và chỉ khi  a = b = c = 1.

Đề không đầy đủ. Bạn xem lại.

\(A=cos10^o.cos50^o.cos70^o=\dfrac{cos10^o}{2}\left[cos120^o+cos20^o\right]=\dfrac{cos10^o}{2}\left[cos20^o-\dfrac{1}{2}\right]\)

\(=\dfrac{1}{2}\left[cos10^o.cos20^o-\dfrac{1}{2}cos10^o\right]\) 

\(=\dfrac{1}{2}\left[\dfrac{1}{2}\left(cos30^o+cos10^o\right)-\dfrac{1}{2}cos10^o\right]=\dfrac{1}{4}.\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{8}\)

\(B=\left(cos\dfrac{\pi}{5}+cos\dfrac{4\pi}{5}\right)+\left(cos\dfrac{2\pi}{5}+cos\dfrac{3\pi}{5}\right)\) 

\(=2cos\dfrac{\pi}{2}.cos\dfrac{3\pi}{10}+2cos\dfrac{\pi}{2}.cos\dfrac{\pi}{10}=0\)

\(C=cos\dfrac{2\pi}{7}+cos\dfrac{4\pi}{7}+cos\dfrac{6\pi}{7}\)   

 \(\Rightarrow C.sin\dfrac{\pi}{7}=sin\dfrac{\pi}{7}.cos\dfrac{2\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{4\pi}{7}+sin\dfrac{\pi}{7}.cos\dfrac{6\pi}{7}\)

\(=\dfrac{1}{2}\left[sin\dfrac{3\pi}{7}+sin\dfrac{-\pi}{7}+sin\dfrac{5\pi}{7}+sin\dfrac{-3\pi}{7}+sin\pi+sin\dfrac{-5\pi}{7}\right]\)

\(=\dfrac{1}{2}\left(sin\pi-sin\dfrac{\pi}{7}\right)=-\dfrac{1}{2}sin\dfrac{\pi}{7}\) \(\Rightarrow C=-\dfrac{1}{2}\)

Mn giúp em giải bài này với ạ.