\(1+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge6\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)\ge2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-1\right)\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}+1\right)\le0\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\le1\)
Ta có:
\(\dfrac{10}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{a}+...+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{12^2}{10a+b+c}\)
Tương tự: \(\dfrac{1}{a}+\dfrac{10}{b}+\dfrac{1}{c}\ge\dfrac{12^2}{a+10b+c}\) ; \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{10}{c}\ge\dfrac{12^2}{a+b+10c}\)
Cộng vế:
\(12^2\sum\dfrac{1}{10a+b+c}\le12\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\le12\)
\(\Rightarrow\sum\dfrac{1}{10a+b+c}\le\dfrac{1}{12}\) (đpcm)
