Question

cho a+b+c=3

c/m \(\dfrac{a}{1+b^2}+\dfrac{b}{1+c^2}+\dfrac{c}{1+a^2}\ge\dfrac{3}{2}\)

Answer 1

Áp dụng bất đẳng thức AM-GM ta có:

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

Chứng minh tương tự ta được:\(\left\{{}\begin{matrix}\dfrac{b}{1+c^2}\ge b-\dfrac{bc}{2}\\\dfrac{c}{1+a^2}\ge c-\dfrac{ac}{2}\end{matrix}\right.\)

Cộng theo vế:

\(L\ge a+b+c-\dfrac{ab+bc+ac}{2}\)

Mặt theo AM-GM: \(ab+bc+ac\le\dfrac{\left(a+b+c\right)^2}{3}=3\)

\(L\ge a+b+c-\dfrac{3}{2}=3-\dfrac{3}{2}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi: \(a=b=c=1\)