a.

Bình phương 2 vế, BĐT cần chứng minh trở thành:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge6\)

Ta có:

\(\sqrt{\left(a^2+1\right)\left(1+b^2\right)}\ge\sqrt{\left(a+b\right)^2}=a+b\)

Tương tự cộng lại:

\(\sqrt{\left(a^2+1\right)\left(b^2+1\right)}+\sqrt{\left(b^2+1\right)\left(c^2+1\right)}+\sqrt{\left(c^2+1\right)\left(a^2+1\right)}\ge2\left(a+b+c\right)=6\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

b.

\(\sum\dfrac{a+1}{a^2+2a+3}=\sum\dfrac{a+1}{a^2+1+2a+2}\le\sum\dfrac{a+1}{4a+2}\)

Nên ta chỉ cần chứng minh:

\(\sum\dfrac{a+1}{4a+2}\le1\Leftrightarrow\sum\dfrac{4a+4}{4a+2}\le4\)

\(\Leftrightarrow\sum\dfrac{1}{2a+1}\ge1\)

Đúng đo: \(\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{2c+1}\ge\dfrac{9}{2\left(a+b+c\right)+3}=1\)

Do \(0\le a;b;c\le2\) 

\(\Rightarrow abc+\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)

\(\Leftrightarrow2\left(ab+bc+ca\right)-4\left(a+b+c\right)+8\ge0\)

\(\Leftrightarrow2\left(ab+bc+ca\right)\ge4\)

\(\Leftrightarrow\left(a+b+c\right)^2-\left(a^2+b^2+c^2\right)\ge4\)

\(\Leftrightarrow9-\left(a^2+b^2+c^2\right)\ge4\)

\(\Leftrightarrow a^2+b^2+c^2\le5\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;1;2\right)\) và các hoán vị

đặt \(S=\frac{a}{4b^2+1}+\frac{b}{4c^2+1}+\frac{c}{4a^2+1}\)

\(=\frac{a^3}{4a^2b^2+a^2}+\frac{b^3}{4b^2c^2+b^2}+\frac{c^3}{4a^2c^2+c^2}\ge\frac{\left(a\sqrt{a}+b\sqrt{b}+c\sqrt{c}\right)^2}{4a^2b^2+4b^2c^2+4c^2a^2+a^2+b^2+c^2}\)

xét hiệu:

1-4(a²b²+b²c²+c²a²)-a²-b²-c²

=2ab+2bc+2ca-4(a²b²+b²c²+c²a²)

=2ab(1-2ab)+2bc(1-2bc)+2ca(1-2ca)

ta có:

\(2ab\le\frac{\left(a+b\right)^2}{2}\le\frac{1}{2};2bc\le\frac{\left(b+c\right)^2}{2}\le\frac{1}{2};2ca\le\frac{\left(c+a\right)^2}{2}\le\frac{1}{2}\)

\(\Rightarrow2ab\left(1-2ab\right);2bc\left(1-2bc\right);2ca\left(1-2ca\right)\ge0\)

\(\Rightarrow1\ge4\left(a^2b^2+b^2c^2+c^2a^2\right)+a^2+b^2+c^2\)

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

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

=>đpcm

dấu"=" xảy ra khi 1 số=1;2 số còn lại =0

Ta có: \(\frac{a}{1+b^2}=\frac{a+ab^2-ab^2}{1+b^2}=\frac{a+ab^2}{1+b^2}-\frac{ab^2}{1+b^2}=a-\frac{ab^2}{1+b^2}\ge a-\frac{ab^2}{2b}=a-\frac{ab}{2}\)

Tương tự ta có: \(\frac{b}{1+c^2}\ge b-\frac{bc}{2}\);  \(\frac{c}{1+a^2}\ge c-\frac{ca}{2}\)

Cộng theo vế 3 BĐT trên,ta được: \(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge3-\frac{1}{2}\left(ab+bc+ca\right)\)

Do \(ab+bc+ca\ge\frac{\left(a+b+c\right)^2}{3}\) (dấu "=" xảy ra khi a = b = c) nên ta có:)

\(\frac{a}{1+b^2}+\frac{b}{1+c^2}+\frac{c}{1+a^2}\ge3-\frac{1}{2}\left(ab+bc+ca\right)\ge3-\frac{1}{2}.\frac{\left(a+b+c\right)^2}{3}=\frac{3}{2}^{\left(đpcm\right)}\)