Ta có \(\dfrac{1}{a+b}+\dfrac{1}{b+c}\ge\dfrac{4}{a+2b+c}\)
\(\Rightarrow VT\ge2\left(\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\right)\left(1\right)\)
Ta cần cm \(\dfrac{1}{2a+b+c}\ge\dfrac{4}{a^2+28}\)
\(\Leftrightarrow a^2+28\ge4a+8b+4c\\ \Leftrightarrow2a^2+b^2+c^2+16-4b-8a-4c\ge0\\ \Leftrightarrow2\left(a-2\right)^2+\left(b-2\right)^2+\left(c-2\right)^2\ge0\left(\text{đúng}\right)\)
\(\Leftrightarrow\sum\dfrac{1}{2a+b+c}\ge\sum\dfrac{8}{a^2+28}\)
Kết hợp \(\left(1\right)\) ta đc đpcm
Dấu \("="\Leftrightarrow a=b=c=2\)
\(a^2+28=a^2+12+16=a^2+a^2+b^2+c^2+16\)
\(=2\left(a^2+4\right)+\left(b^2+4\right)+\left(c^2+4\right)\ge2.4a+4b+4c=4\left(2a+b+c\right)\)
\(\Rightarrow\dfrac{8}{a^2+28}\le\dfrac{8}{4\left(2a+b+c\right)}=\dfrac{2}{a+b+a+c}\le\dfrac{1}{4}\left(\dfrac{2}{a+b}+\dfrac{2}{a+c}\right)=\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\)
Tương tự và cộng lại:
\(\dfrac{8}{a^2+28}+\dfrac{8}{b^2+28}+\dfrac{8}{c^2+28}\le\dfrac{1}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}+\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\)
\(\Leftrightarrowđpcm\)
