Question

Cho a, b > 0 và a + b = 1. Tìm GTNN của:

\(A=\dfrac{1}{1+3ab+a^2}+\dfrac{1}{1+3ab+b^2}\)

Answer 1

Áp dụng bđt Cauchy-Schwarz\(A=\dfrac{1}{1+3ab+a^2}+\dfrac{1}{1+3ab+b^2}\)

\(A=\dfrac{1}{1+2ab+ab+a^2}+\dfrac{1}{1+3ab+b^2}\)

\(A\ge\dfrac{\left(1+1\right)^2}{1+2ab+ab+a^2+1+3ab+b^2}\)

\(A\ge\dfrac{4}{\left(a+b\right)^2+4ab+2}=\dfrac{4}{3+4ab}\)

Mặt khác theo AM-GM: \(4ab\le\left(a+b\right)^2\)

\(\Rightarrow\dfrac{4}{3+4ab}\ge\dfrac{4}{3+\left(a+b\right)^2}=\dfrac{4}{3+1}=1\)

\(\Rightarrow A\ge1\)

Dấu "=" xảy ra khi: \(a=b=\dfrac{1}{2}\)