﻿ Given abc=1 and a,b,c>0 Prove 1/a+1/b+1/c+3/(a+b+c) ≥ 4 or \dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{3}{a+b+c}≥4

#### Question

Given abc=1 and a,b,c>0 Prove 1/a+1/b+1/c+3/(a+b+c) ≥ 4

or

\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{3}{a+b+c}≥4

Collected in the board: Inequality

Steven Zheng posted 2 years ago

Define A = \dfrac{1}{a}, B = \dfrac{1}{b},C = \dfrac{1}{c}

\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{3}{a+b+c} is converted to,

A+B+C+\dfrac{3}{AB+BC+AC}
(1)

\because (A-B)^2+(B-C)^2+(A-C)^2≥0

\therefore A^2+B^2+C^2≥AB+BC+AC

\therefore (A+B+C)^2≥3(AB+BC+AC)

AB+BC+AC≤\dfrac{(A+B+C)^2}{3}
(2)

Therefore, the expression (1) satisfies the following

inequality expression

A+B+C+\dfrac{3}{AB+BC+AC}

≤ A+B+C+\dfrac{9}{(A+B+C)^2}
(3)

Define x = A+B+C , expression (3) is converted to the function

f(x)=x+\dfrac{9}{x^2}
(4)

Using the inequality of arithmetic and geometric means

\dfrac{A+B+C}{3}≥ \sqrt[3]{ABC}=1

\therefore x = A+B+C ≥ 3

When x = 3 , f(x) has its minimum value

f(3)=3 + \dfrac{9}{3^2}=4

Now we have completed the proof

\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{3}{a+b+c}≥4

Steven Zheng posted 2 years ago

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