\(3-B=\left(a-\frac{a}{1+b^2}\right)+\left(b-\frac{b}{1+c^2}\right)+\left(c-\frac{c}{1+a^2}\right)=\frac{b^2}{1+b^2}+\frac{c^2}{1+c^2}+\frac{a^2}{1+a^2}\le\frac{b^2}{2b}+\frac{c^2}{2c}+\frac{a^2}{2a}=\frac{1}{2}\left(a+b+c\right)=\frac{3}{2}\)
=> \(B\ge\frac{3}{2}\)
Dấu "=" xảy ra <=> a = b = c = 1
\(B=\frac{a\left(b^2+1\right)-ab^2}{b^2+1}+\frac{b\left(c^2+1\right)-bc^2}{c^2+1}+\frac{c\left(a^2+1\right)-ca^2}{c^2+1}\)
\(\Leftrightarrow B=a-\frac{ab^2}{b^2+1}+b-\frac{bc^2}{c^2+1}+c-\frac{ca^2}{a^2+1}\)
\(\Leftrightarrow B=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\right)\)
+ \(b^2+1\ge2b\forall b\)
\(\Rightarrow\frac{ab^2}{b^2+1}\le\frac{ab^2}{2b}=\frac{ab}{2}\). Dấu "=" xảy ra \(\Leftrightarrow b=1\)
+ Tương tự ta cm đc :
\(\frac{bc^2}{c^2+1}\le\frac{bc}{2}\) . Dấu "=" xảy ra \(\Leftrightarrow c=1\)
\(\frac{ca^2}{a^2+1}\le\frac{ca}{2}\). Dấu '=" xảy ra \(\Leftrightarrow a=1\)
Do đó : \(\frac{ab^2}{a^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\le\frac{ab+bc+ca}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
+ \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\forall a,b,c\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)\ge0\)
\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(a+bc+ca\right)\)
\(\Leftrightarrow ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}=3\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
Do đó : \(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\le\frac{ab+bc+ca}{2}\le\frac{3}{2}\)
\(\Leftrightarrow-\left(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\right)\ge-\frac{3}{2}\)
\(\Leftrightarrow B=\left(a+b+c\right)-\left(\frac{ab^2}{b^2+1}+\frac{bc^2}{c^2+1}+\frac{ca^2}{a^2+1}\right)\)
\(\ge3-\frac{3}{2}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
