\(\dfrac{a}{1+b^2}=a-\dfrac{ab^2}{1+b^2}\ge a-\dfrac{ab^2}{2b}=a-\dfrac{1}{2}ab\)

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

Cộng vế:

\(P\ge a+b+c-\dfrac{1}{2}\left(ab+bc+ca\right)\ge a+b+c-\dfrac{1}{6}\left(a+b+c\right)^2=\dfrac{3}{2}\)

\(P_{min}=\dfrac{3}{2}\) khi \(a=b=c=1\)

\(S=\dfrac{1}{a^3+b^3}+\dfrac{\dfrac{9}{4}}{3a^2b}+\dfrac{\dfrac{9}{4}}{3ab^2}+\dfrac{1}{4ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Áp dụng bđt Cauchy-Schwarz dạng Engel có:

\(S\ge\dfrac{\left(1+\dfrac{3}{2}+\dfrac{3}{2}\right)^2}{a^3+3a^2b+3ab^2+b^3}+\dfrac{1}{4ab}.\dfrac{4}{a+b}\)

\(\Leftrightarrow S\ge\dfrac{16}{\left(a+b\right)^3}+\dfrac{1}{\left(a+b\right)^2}.\dfrac{4}{a+b}\)

\(\Leftrightarrow S\ge\dfrac{16}{1}+\dfrac{1}{1}.\dfrac{4}{1}=20\)

Dấu "=" xảy ra khi \(a=b=\dfrac{1}{2}\)

Vậy GTNN của \(S=20\) khi \(a=b=\dfrac{1}{2}\)

Lời giải:
Áp dụng BĐT Cô-si:

$a^2+1\geq 2a$

$b^2+4\geq 4b$

$\Rightarrow a^2+b^2\geq 2a+4b-5$

$\Rightarrow P\geq 2a+4b-5+\frac{1}{a+b}+\frac{1}{b}$

$=\frac{a+b}{9}+\frac{1}{a+b}+(\frac{b}{4}+\frac{1}{b})+\frac{17}{9}a+\frac{131}{36}b-5$

$\geq 2\sqrt{\frac{1}{9}}+2\sqrt{\frac{1}{4}}+\frac{17}{9}a+\frac{131}{36}b-5$

$=\frac{2}{3}+1+\frac{17}{9}a+\frac{131}{36}b-5$

$\geq \frac{2}{3}+1+\frac{17}{9}+\frac{131}{36}.2-5=\frac{35}{6}$

Vậy $P_{\min}=\frac{35}{6}$ khi $a=1; b=2$

\(GT\Rightarrow a+b=5\)

\(P=\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}=\dfrac{4}{5}\)

Dấu "=" xảy ra khi \(a=b=\dfrac{5}{2}\)

Ta có BĐT: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)=3.3=9\)

\(\Rightarrow a+b+c\ge3\)

Phân tích và áp dụng BĐT AM-GM:

\(\dfrac{1+3a}{1+b^2}=\dfrac{1}{1+b^2}+\dfrac{3a}{1+b^2}=\left(1-\dfrac{b^2}{1+b^2}\right)+\left(3a-\dfrac{3ab^2}{1+b^2}\right)\ge\left(1-\dfrac{b^2}{2b}\right)+\left(3a-\dfrac{3ab^2}{2b}\right)=\left(1-\dfrac{b}{2}\right)+\left(3a-\dfrac{3}{2}ab\right)\)

Tương tự:

\(\dfrac{1+3b}{1+c^2}\ge\left(1-\dfrac{c}{2}\right)+\left(3b-\dfrac{3}{2}bc\right)\)

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

Cộng các vế của các BĐT ta được:

\(P\ge3-\dfrac{1}{2}\left(a+b+c\right)+3\left(a+b+c\right)-\dfrac{3}{2}\left(ab+bc+ca\right)=3+\dfrac{5}{2}\left(a+b+c\right)-\dfrac{3}{2}.3\ge3+\dfrac{5}{2}.3-\dfrac{9}{2}=6\)

\(P=6\Leftrightarrow a=b=c=1\)

Vậy \(P_{min}=6\)