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

\(P\ge\dfrac{1}{2}\left(a+b\right)^2+\dfrac{8}{\left(a+b\right)^2}=\dfrac{1}{2}\left(a+b\right)^2+\dfrac{1}{2\left(a+b\right)^2}+\dfrac{15}{2\left(a+b\right)^2}\)

\(P\ge\dfrac{1}{2}.2\sqrt{\dfrac{\left(a+b\right)^2}{\left(a+b\right)^2}}+\dfrac{15}{2.1^2}=\dfrac{17}{2}\)

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

\(P=\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}+\dfrac{1}{a^2+b^2+c^2}\ge\dfrac{\left(1+1+1\right)^2}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\) (BĐT Cauchy Schwarz)

\(=\dfrac{9}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}\)

\(=\dfrac{1}{ab+bc+ca}+\dfrac{1}{ab+bc+ca}+\dfrac{1}{a^2+b^2+c^2}+\dfrac{7}{ab+bc+ca}\)

\(\ge\dfrac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2ac+2bc}+\dfrac{7}{ab+bc+ca}\)

\(=\dfrac{9}{\left(a+b+c\right)^2}+\dfrac{7}{ab+bc+ca}\)

Ta có: \(ab+bc+ca\le\dfrac{\left(a+b+c\right)^2}{3}=\dfrac{1}{3}\) .Thế vào biểu thức

\(\Rightarrow P\ge9+\dfrac{7}{\dfrac{1}{3}}=9+21=30\)

\(\Rightarrow P_{min}=30\) khi \(a=b=c=\dfrac{1}{3}\)

\(S\ge0\), đẳng thức xảy ra  khi a = b = 0.

Bài này chắc có vấn đề, đáng lẽ phải là tìm GTLN

Lời giải:

$a^2+b^2=a+b$

$\Rightarrow (a+b)^2-(a+b)=2ab\geq 0$

$\Rightarrow a+b\geq 1$. Do đó:

$S=\frac{a}{a+1}+\frac{b}{b+1}=\frac{2ab+a+b}{ab+a+b+1}\geq \frac{\frac{ab}{2}+\frac{a+b+1}{2}}{ab+a+b+1}=\frac{1}{2}$
Vậy GTNN của $S$ là $\frac{1}{2}$. Dấu "=" xảy ra khi $(a,b)=(0,1)$ và hoán vị.

Đây là bài IMO 2001 và không cần điều kiện \(a+b+c=1\)

Áp dụng Holder:

\(P.P.\left[a\left(a^2+8bc\right)+b\left(b^2+8ac\right)+c\left(c^2+8ab\right)\right]\ge\left(a+b+c\right)^3\)

\(\Leftrightarrow P^2\ge\dfrac{\left(a+b+c\right)^3}{a^3+b^3+c^3+24abc}=\dfrac{a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)}{a^3+b^3+c^3+24abc}\)

\(\Rightarrow P^2\ge\dfrac{a^3+b^3+c^3+3.2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}}{a^3+b^3+c^3+24abc}=1\)

\(\Rightarrow P\ge1\)

\(A=\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\left(ab+\dfrac{16}{ab}\right)+\dfrac{17}{2ab}\)

\(A\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{16ab}{ab}}+\dfrac{17}{\dfrac{2\left(a+b\right)^2}{4}}\)

\(A\ge\dfrac{4}{\left(a+b\right)^2}+8+\dfrac{34}{\left(a+b\right)^2}\ge\dfrac{4}{4^2}+8+\dfrac{34}{4^2}=\dfrac{83}{8}\)

Dấu "=" xảy ra khi \(a=b=2\)