﻿ 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}

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

Collected in the board: Inequality

Steven Zheng posted 2 months ago

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.

Steven Zheng posted 2 months ago

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