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
Answer
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