Áp dụng BĐT AM - GM ta có:

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

Chứng minh tương tự ta có:

\(\frac{b+1}{c^2+1}\ge b+1-\frac{c+bc}{2}\left(2\right)\)

\(\frac{c+1}{a^2+1}\ge c+1-\frac{a+ca}{2}\left(3\right)\)

Từ: \(\left(1\right)\left(2\right)\left(3\right)\Rightarrow\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\ge\frac{a+b+c}{2}+3-\frac{ab+bc+ca}{2}\)

Lại có: \(a^2+b^2+c^2\ge ab+bc+ca\)

Hay: \(3\left(ab+bc+ac\right)\le\left(a+b+c\right)^2=3^2=9\)

Vì vậy: \(\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\ge\frac{a+b+c}{2}+3-\frac{ab+bc+ca}{2}=\frac{3}{2}+3-\frac{9}{6}=3\)

\(\Rightarrow\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\ge3\)

\(\Rightarrow Min_P=3\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c=1\)

* Dũng kỹ thuật Cô-si ngược dấu

\(P=\left(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}\right)+\left(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\right)\)

+ \(\frac{a}{b^2+1}+\frac{b}{c^2+1}+\frac{c}{a^2+1}=a-\frac{ab^2}{b^2+1}+b-\frac{bc^2}{c^2+1}+c-\frac{ca^2}{a^2+1}\)

\(\ge3-\left(\frac{ab^2}{2b}+\frac{bc^2}{2c}+\frac{ca^2}{2a}\right)=3-\frac{ab+bc+ca}{2}\ge3-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=\frac{3}{2}\)

Dấu "=" \(\Leftrightarrow a=b=c=1\)

+ \(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}=1-\frac{a^2}{a^2+1}+1-\frac{b^2}{b^2+1}+1-\frac{c^2}{c^2+1}\ge3-\left(\frac{a^2}{2a}+\frac{b^2}{2b}+\frac{c^2}{2c}\right)=3-\frac{a+b+c}{2}=\frac{3}{2}\)

Dấu "=" \(\Leftrightarrow a=b=c=1\)

Do đó: \(P\ge3\). Dấu "=" \(\Leftrightarrow a=b=c=1\)

Câu a : Áp dụng BĐT Cô - si cho các số dương ta có :

\(\left\{{}\begin{matrix}\frac{a^2}{b}+b\ge2\sqrt{\frac{a^2}{b}.b}=2a\\\frac{b^2}{c}+c\ge2\sqrt{\frac{b^2}{c}.c}=2b\\\frac{c^2}{a}+a\ge2\sqrt{\frac{c^2}{a}.a}=2c\end{matrix}\right.\)

Cộng từng vế của BĐT ta thu được :

\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge2a+2b+c\)

\(\Leftrightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c=1\) ( đpcm )

Dấu \("="\) xảy ra khi \(a=b=c=\frac{1}{3}\)

\(A=\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\)

\(A=\Sigma\left(a-\frac{ab^2}{1+b^2}\right)\)

Áp dụng bất đẳng thức Cô-si :

\(A\ge\Sigma\left(a-\frac{ab^2}{2b}\right)=\Sigma\left(a-\frac{ab}{2}\right)\)

\(=\left(a+b+c\right)-\left(\frac{ab+bc+ca}{2}\right)\)\(\ge3-\frac{\frac{\left(a+b+c\right)^2}{3}}{2}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

Áp dụng bất đẳng thức Bunhiacopxki :

\(\left(1^2+4^2\right)\left(a^2+\frac{1}{b^2}\right)\ge\left(a+\frac{4}{b}\right)^2\)

\(\Leftrightarrow17\cdot\left(a^2+\frac{1}{b^2}\right)\ge\left(a+\frac{4}{b}\right)^2\)

\(\Leftrightarrow\sqrt{17}\cdot\sqrt{a^2+\frac{1}{b^2}}\ge a+\frac{4}{b}\)

Tương tự ta có :

\(\sqrt{17}\cdot\sqrt{b^2+\frac{1}{c^2}}\ge b+\frac{4}{c}\)

\(\sqrt{17}\cdot\sqrt{c^2+\frac{1}{a^2}}\ge c+\frac{4}{a}\)

Cộng theo vế của 3 bđt ta được :

\(\sqrt{17}\cdot\left(\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}\right)\ge a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)

\(\Leftrightarrow\sqrt{17}\cdot A\ge a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)

Áp dụng bất đẳng thức Cô-si :

\(a+b+c+\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\)

\(=16a+\frac{4}{a}+16b+\frac{4}{b}+16c+\frac{4}{c}-15a-15b-15c\)

\(\ge2\sqrt{\frac{4\cdot16a}{a}}+2\sqrt{\frac{4\cdot16b}{b}}+2\sqrt{\frac{4\cdot16c}{c}}-15\left(a+b+c\right)\)

\(\ge16+16+16-15\cdot\frac{3}{2}=\frac{51}{2}\)

Do đó : \(\sqrt{17}\cdot A\ge\frac{51}{2}\)

\(\Leftrightarrow A\ge\frac{3\sqrt{17}}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{2}\)