\(\frac{ab}{c+1}=\frac{ab}{a+c+b+c}\le\frac{ab}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)=\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)

\(M\le\frac{1}{4}\left[\frac{ab}{a+c}+\frac{ab}{b+c}+\frac{bc}{a+b}+\frac{bc}{c+a}+\frac{ca}{a+b}+\frac{ca}{b+c}\right]\)

\(=\frac{1}{4}\left[\frac{b\left(a+c\right)}{a+c}+\frac{a\left(b+c\right)}{b+c}+\frac{c\left(a+b\right)}{a+b}\right]=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\)

 chịu thôi chị ơi!

Ai trả lời câu này được bái luôn thành sư phụ!!!!!

cảm ơn nha, giờ chị giải được rồi 

Ta dễ có bất đẳng thức \(a^2+b^2+c^2\ge ab+bc+ca\)

Dấu bằng xảy ra khi a = b = c

Mà a+b+c=2021 nên a=b=c=2021/3

\(\frac{a^2}{a+bc}=\frac{a^3}{a^2+abc}=\frac{a^3}{a^2+ab+bc+ac}=\frac{a^3}{\left(a+b\right)\left(a+c\right)}\)

Áp dụng BĐT cosi

\(\frac{a^3}{\left(a+b\right)\left(a+c\right)}+\frac{a+b}{8}+\frac{a+c}{8}\ge\frac{3}{4}a\)

Tương tự 

=> \(A\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{2}\left(a+b+c\right)=\frac{1}{4}\left(a+b+c\right)\)

Lại có \(\left(a+b+c\right)\ge\frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\frac{9}{1}=9\)

=> \(A\ge\frac{9}{4}\)

MinA=9/4 khi a=b=c=3

ta có:

\(abc=ab+bc+ca\Rightarrow1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Lại có:

\(\frac{a^2}{b^3}+\frac{1}{a}+\frac{1}{a}\ge\frac{3}{b},\frac{b^2}{c^3}+\frac{1}{b}+\frac{1}{b}\ge\frac{3}{c},\frac{c^2}{a^3}+\frac{1}{c}+\frac{1}{c}\ge\frac{3}{a}\)

\(\Rightarrow P+2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\Rightarrow P\ge\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)

Ta có \(a+bc=a\left(a+b+c\right)+bc=\left(a+b\right)\left(a+c\right)\)

        \(b+ac=\left(b+a\right)\left(b+c\right)\)

        \(c+ab=\left(a+b\right)\left(c+b\right)\)

Đặt \(a+b=x;b+c=y;a+c=z\)=> \(x+y+z=2\)

Khi đó \(P=\frac{xy}{z}+\frac{yz}{x}+\frac{xz}{y}\)

Áp dụng BĐT cosi \(\frac{xy}{z}+\frac{yz}{x}\ge2y\); \(\frac{yz}{x}+\frac{xz}{y}\ge2z\);\(\frac{xy}{z}+\frac{xz}{y}\ge2z\)

Cộng 3 BĐT trên

=> \(P\ge x+y+z=2\)

Vậy MinP=2 khi a=b=c=1/3

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

\(P_{min}=1\) khi \(a=b=c=1\)

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

Do \(a;b\ge1\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\Rightarrow ab\ge a+b-1=2-c\)

\(\Rightarrow ab+c\left(a+b\right)\ge2-c+c\left(3-c\right)=-c^2+2c+2=c\left(2-c\right)+2\ge2\)

\(\Rightarrow P\le\dfrac{9}{2}-2=\dfrac{5}{2}\)

\(P_{max}=\dfrac{5}{2}\) khi \(\left(a;b;c\right)=\left(1;2;0\right);\left(2;1;0\right)\)