\(VT=\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\)
\(=1-\frac{a^2}{a^2+1}+1-\frac{b^2}{b^2+1}+1-\frac{c^2}{c^2+1}\)
\(=3-\left(\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\right)\)
Áp dụng bất đẳng thức Cauchy :
\(VT\ge3-\left(\frac{a^2}{2a}+\frac{b^2}{2b}+\frac{c^2}{2c}\right)=3-\left(\frac{a}{2}+\frac{b}{2}+\frac{c}{2}\right)\)
\(=3-\frac{a+b+c}{2}=3-\frac{3}{2}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)
\(ab+ac+bc\le a^2+b^2+c^2\\ \Rightarrow3\left(ab+ac+bc\right)\le a^2+b^2+c^2+2\left(ab+ac+bc\right)\\ \Rightarrow3\left(ab+ac+bc\right)\le\left(a+b+c\right)^2=9\\ \Rightarrow ab+ac+bc\le3\\ \Rightarrow2\left(ab+ac+bc\right)\le6\)
Áp dụng BDT Cô-si với 3 số dương:
\(\Rightarrow\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\ge\frac{9}{a^2+1+b^2+1+c^2+1}\\ =\frac{9}{a^2+b^2+c^2+3}=\frac{9}{a^2+b^2+c^2+6-3}\\ \ge\frac{9}{a^2+b^2+c^2+2\left(ab+ac+bc\right)-3}=\frac{9}{\left(a+b+c\right)^2-3}=\frac{9}{6}=\frac{3}{2}\)
Dấu "=" xảy ra khi a=b=c=1
