Từ \(a+b+c=abc\Rightarrow\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a}{b^3}+\dfrac{1}{ab}\ge\dfrac{2}{b^2}\). Tương tự cho 2 BĐT trên ta cũng có:
\(\dfrac{b}{c^3}+\dfrac{1}{bc}\ge\dfrac{2}{c^2};\dfrac{c}{a^3}+\dfrac{1}{ca}\ge\dfrac{1}{a^2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge\dfrac{2}{a^2}+\dfrac{2}{b^2}+\dfrac{2}{c^2}\)
Lại có BĐT quen thuộc \(\dfrac{2}{a^2}+\dfrac{2}{b^2}+\dfrac{2}{c^2}\ge\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\)
\(\Rightarrow VT+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}\ge\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ca}\)
\(\Rightarrow VT\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=1=VP\)
Đẳng thức xảy ra khi \(a=b=c=\sqrt{3}\)
