Question

Cho a;b;c >0 và a+b+c=3

Tìm Min P=\(\dfrac{a+3}{3a+bc}+\dfrac{b+3}{3b+ca}+\dfrac{c+3}{3c+ab}.\)

Answer 1

3a + bc = a(a + b + c) + bc = a² + ab + ac + bc = a(a + b) + c(a + b)

= (a + b)(a + c)

\(\dfrac{a+3}{3a+bc}=\dfrac{\left(a+b\right)+\left(a+c\right)}{\left(a+b\right)\left(a+c\right)}=\dfrac{1}{a+c}+\dfrac{1}{a+b}\)

Tương tự, ta có:

\(\dfrac{b+3}{3b+ac}=\dfrac{1}{b+c}+\dfrac{1}{a+b}\)

\(\dfrac{c+3}{3c+ab}=\dfrac{1}{a+c}+\dfrac{1}{b+c}\)

Áp dụng BĐT Cauchy Schwarz dạng Engel, ta có:

\(P=\dfrac{a+3}{3a+bc}+\dfrac{b+3}{3b+ac}+\dfrac{c+3}{3c+ab}\)

\(=\dfrac{1}{a+c}+\dfrac{1}{a+b}+\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{b+c}\)

\(=2\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{a+c}\right)\)

\(\ge2\left[\dfrac{\left(1+1+1\right)^2}{2\left(a+b+c\right)}\right]=2\times\dfrac{9}{2\times3}=3\)

Dấu "=" xảy ra khi a = b = c = 1

Vậy Min P = 3 <=> a = b = c = 1