Ta có:
\(\frac{a}{1+b^2}=a-\frac{ab^2}{1+b^2}\)
Áp dụng BĐT AM-GM cho các số không âm, ta có:
\(1+b^2\ge2b\Rightarrow\frac{1}{1+b^2}\le\frac{1}{2b}\Rightarrow-\frac{1}{1+b^2}\ge-\frac{1}{2b}\)\(\Rightarrow-\frac{ab^2}{1+b^2}\ge-\frac{ab}{2}\)
\(\Rightarrow\frac{a}{1+b^2}\ge a-\frac{ab}{2}\)
CMTT: \(\frac{b}{1+c^2}\ge b-\frac{bc}{2};\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)
\(\Rightarrow BĐT\ge a+b+c-\frac{ab+bc+ca}{2}\)\(=3-\frac{ab+bc+ca}{2}\)
Mặt khác ta có:
\(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)
\(\Leftrightarrow3\ge ab+bc+ca\)\(\Leftrightarrow-\frac{3}{2}\le-\frac{ab+bc+ca}{2}\)
\(\Rightarrow BĐT\ge3-\frac{3}{2}=\frac{3}{2}\)(đpcm)
\(''=''\Leftrightarrow a=b=c=1\)
