Question

cho a,b,c  thỏa mãn a+b+c=3. cmr :

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

Answer 1

a;b;c phải là số dương chứ bạn?

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

Tương tự:

\(\dfrac{b+1}{c^2+1}\ge b+1-\dfrac{c+bc}{2}\) ; \(\dfrac{c+1}{a^2+1}\ge c+1-\dfrac{a+ca}{2}\)

Cộng vế với vế:

\(VT\ge a+b+c+3-\dfrac{1}{2}\left(a+b+c+ab+bc+ca\right)\)

\(VT\ge6-\dfrac{3}{2}-\dfrac{1}{2}\left(ab+bc+ca\right)\ge\dfrac{9}{2}-\dfrac{1}{6}\left(a+b+c\right)^2=3=a+b+c\)

\(\Rightarrow VT\ge a+b+c\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)