Question
Let a,b,c be real numbers such that ab+bc+ac=1. Prove the inequality
\Big( a+\dfrac{1}{b} \Big) ^2+\Big( b+\dfrac{1}{c} \Big) ^2+\Big( c+\dfrac{1}{a} \Big) ^2\geq 16
Answer
Expanding the LHS gives
a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} +2\Big( \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\Big)
(1)
Given ab+bc+ac=1,
1 = ab+bc+ac\geq 3\sqrt[3]{a^2b^2c^2}
Therefore,
\dfrac{1}{\sqrt[3]{a^2b^2c^2} } \geq 3
(2)
Now Let's evaluate three parts of expression (1)
The first part
a^2+b^2+c^2
=a^2+b^2+c^2-1+1
=a^2+b^2+c^2- ab+bc+ac+1
=\dfrac{1}{2}[(a-b)^2+(b-c)^2+(a-c)^2] +1
\geq 1
The second part
\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}
\geq 3\sqrt[3]{\dfrac{1}{a^2b^2c^2} }
\geq 3\cdot 3 = 9
The third part
2\Big( \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\Big)
\geq 2\cdot 3\sqrt[3]{\dfrac{a}{b}\cdot \dfrac{b}{c}\cdot \dfrac{c}{a} }
=6
Combine the conclusion of the three parts. We get
a^2+b^2+c^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} +2\Big( \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\Big)
\geq 1+9+6 =16
Now we have proved the inequality.