\(\dfrac{a+1}{b^2+1}=a+1-\dfrac{ab^2+b^2}{b^2+1}\) minh hong biet phai lam gi tiep theo dau
Áp dụng BĐT Cauchy , ta có :
\(\dfrac{a+1}{b^2+1}=a+1-\dfrac{b^2\left(a+1\right)}{b^2+1}\ge a+1-\dfrac{b^2\left(a+1\right)}{2b}=a+1-\dfrac{ab+b}{2}\)
\(\dfrac{b+1}{c^2+1}=b+1-\dfrac{c^2\left(b+1\right)}{c^2+1}\ge b+1-\dfrac{c^2\left(b+1\right)}{2c}=b+1-\dfrac{bc+c}{2}\)
\(\dfrac{c+1}{a^2+1}=c+1-\dfrac{a^2\left(c+1\right)}{a^2+1}\ge c+1-\dfrac{a^2\left(c+1\right)}{2a}=c+1-\dfrac{ac+a}{2}\)
\(\Rightarrow\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\ge a+1+b+1+c+1-\dfrac{ab+b}{2}-\dfrac{bc+c}{2}-\dfrac{ac+a}{2}=\dfrac{9-ab-bc-ac}{2}\ge\dfrac{9-3}{2}=3\)
\("="\Leftrightarrow x=y=1\)
Đặt \(A=\dfrac{a+1}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\)
\(3-A=\left(1-\dfrac{a+1}{b^2+1}\right)+\left(1-\dfrac{b+1}{c^2+1}\right)+\left(1-\dfrac{c+1}{a^2+1}\right)\)
\(3-A=\dfrac{b^2-a}{b^2+1}+\dfrac{b+1}{c^2+1}+\dfrac{c+1}{a^2+1}\)
\(3-A\le\dfrac{b^2-a}{2b}+\dfrac{c^2-b}{2c}+\dfrac{a^2-c}{2a}\) (Bđt AM-GM)
\(6-2A\le\left(a+b+c\right)-\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\le3-3=0\)
=> \(A\ge3\)
Dấu "=" xảy ra <=> a = b = c
