Question

Tìm giá trị nhỏ nhất

\(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)

với a,b,c>0

Answer 1

Lời giải:

Áp dụng BĐT Cauchy-Schwarz:

\(A=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ba}+\frac{c^2}{ca+cb}\)

\(\geq \frac{(a+b+c)^2}{2(ab+bc+ac)}\)

Mà: \(\frac{(a+b+c)^2}{2(ab+bc+ac)}-\frac{3}{2}=\frac{1}{2}.\frac{(a+b+c)^2-3(ab+bc+ac)}{ab+bc+ac}=\frac{1}{2}.\frac{a^2+b^2+c^2-ab-bc-ac}{2(ab+bc+ac)}\)

\(=\frac{1}{4}.\frac{(a-b)^2+(b-c)^2+(c-a)^2}{ab+bc+ac}\geq 0, \forall a,b,c>0\)

Do đó: \(A\geq \frac{(a+b+c)^2}{2(ab+bc+ac)}\geq \frac{3}{2}\)

Vậy $A_{\min}=\frac{3}{2}$ khi $a=b=c$