Biến đổi like that:
\(VT=\sum\dfrac{1}{ab}=\sum\dfrac{ab+bc+ca}{ab}=\sum\left(\dfrac{c}{a}+\dfrac{c}{b}\right)+3\)
nên chỉ cần :\(\sum\left(\dfrac{c}{a}+\dfrac{c}{b}\right)\ge\sum\sqrt{\dfrac{1}{a^2}+1}=\sum\dfrac{\sqrt{a^2+1}}{a}\)
Áp dụng AM-GM:
\(\dfrac{\sqrt{a^2+1}}{a}=\dfrac{\sqrt{a^2+ab+bc+ca}}{a}=\dfrac{\sqrt{\left(a+b\right)\left(a+c\right)}}{a}\le\dfrac{2a+b+c}{2a}\)
Áp dụng tương tự:
\(VP=\sum\dfrac{\sqrt{a^2+1}}{a}\le\sum\dfrac{2a+b+c}{2a}=3+\dfrac{1}{2}\sum\left(\dfrac{b}{a}+\dfrac{c}{a}\right)\)
BĐT đúng khi ta chứng minh được
\(VT=\sum\left(\dfrac{c}{a}+\dfrac{c}{b}\right)\ge3+\dfrac{1}{2}\sum\left(\dfrac{c}{a}+\dfrac{c}{b}\right)\)
Điều này hiển nhiên đúng theo AM-GM:
\(\dfrac{1}{2}\sum\left(\dfrac{c}{a}+\dfrac{c}{b}\right)=\dfrac{1}{2}\left(\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{b}{c}\right)\ge\dfrac{1}{2}.6\sqrt[6]{\dfrac{a^2b^2c^2}{a^2b^2c^2}}=3\)
\(\Rightarrow\)đpcm
Dấu = xảy ra khi a=b=c
