\(\dfrac{b+c-a}{2a}+\dfrac{a-b+c}{2b}+\dfrac{a+b-c}{2c}\ge\dfrac{3}{2}\)
Ta có: \(\dfrac{b+c-a}{2a}=\dfrac{b}{2a}+\dfrac{c}{2a}-\dfrac{a}{2a}=\dfrac{b}{2a}+\dfrac{c}{2a}-\dfrac{1}{2}\)
Viết lại BĐT cần chứng minh như sau:
\(\dfrac{b}{2a}+\dfrac{c}{2a}-\dfrac{1}{2}+\dfrac{a}{2b}-\dfrac{1}{2}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-\dfrac{1}{2}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-\dfrac{3}{2}\ge\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-3\ge0\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{b}{2a}+\dfrac{a}{2b}=\dfrac{1}{2}\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{b}{a}\cdot\dfrac{a}{b}}=2\cdot\dfrac{1}{2}=1\)
\(\dfrac{c}{2a}+\dfrac{a}{2c}=\dfrac{1}{2}\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{c}{a}+\dfrac{a}{c}}=\dfrac{1}{2}\cdot2=1\)
\(\dfrac{b}{2c}+\dfrac{c}{2b}=\dfrac{1}{2}\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge\dfrac{1}{2}\cdot2\sqrt{\dfrac{b}{c}\cdot\dfrac{c}{b}}=\dfrac{1}{2}\cdot2=1\)
Cộng theo vế 3 BĐT trên ta có:
\(\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}\ge3\)
\(\Rightarrow\dfrac{b}{2a}+\dfrac{c}{2a}+\dfrac{a}{2b}+\dfrac{c}{2b}+\dfrac{a}{2c}+\dfrac{b}{2c}-3\ge3-3=0\)
BĐT đúng nên ta có ĐPCM
