Lời giải:

Áp dụng BĐT AM-GM:

$1\geq a+b\geq 2\sqrt{ab}\Rightarrow ab\leq \frac{1}{4}$

Áp dụng BĐT Cauchy-Schwarz:
\(A=\frac{1}{1+a^2+b^2}+\frac{1}{6ab}+\frac{1}{3ab}\geq \frac{4}{1+a^2+b^2+6ab}+\frac{1}{3ab}\)

\(=\frac{4}{1+(a+b)^2+4ab}+\frac{1}{3ab}\geq \frac{4}{1+1+4.\frac{1}{4}}+\frac{1}{3.\frac{1}{4}}=\frac{8}{3}\)

Vậy $A_{\min}=\frac{8}{3}$ khi $a=b=\frac{1}{2}$

B1 

Ta có

\(A=\frac{a^2}{24}+\frac{9}{a}+\frac{9}{a}+\frac{23a^2}{24}\ge3\sqrt[3]{\frac{a^2}{24}.\frac{9}{a}.\frac{9}{a}+\frac{23a^2}{24}}\ge\frac{9}{2}+\frac{23.36}{24}\ge39\)

Dấu "=" xảy ra <=> a=6

Vậy Min A = 39 <=> a=6

 \(A=a^2+\frac{18}{a}=a^2+\frac{216}{a}+\frac{216}{a}-\frac{414}{a}\ge3\sqrt[3]{a^2.\frac{216}{a}.\frac{216}{a}}-69=39\)

Đẳng thức xảy ra khi a = 6

B3: Áp dụng bđt AM-GM

\(A=\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}+\frac{3\left(a+b\right)}{4\sqrt{ab}}\ge2\sqrt{\frac{a+b}{4\sqrt{ab}}.\frac{\sqrt{ab}}{a+b}}+\frac{3\left(a+b\right)}{4\left(\frac{a+b}{2}\right)}\)

\(=1+\frac{3\left(a+b\right)}{2\left(a+b\right)}=1+\frac{3}{2}=\frac{5}{2}\)

Dấu "=" xảy ra khi \(a=b>0\)

Dự đoán điểm rơi \(a=b=c=\frac{1}{3}\)

Khi đó:

\(S=a+b+c+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

\(=\left(a+b+c+\frac{1}{9a}+\frac{1}{9b}+\frac{1}{9c}\right)+8\left(\frac{1}{9a}+\frac{1}{9b}+\frac{1}{9c}\right)\)

\(\ge6\sqrt[6]{a\cdot b\cdot c\cdot\frac{1}{9a}\cdot\frac{1}{9b}\cdot\frac{1}{9c}}+24\sqrt[3]{\frac{1}{9a}\cdot\frac{1}{9b}\cdot\frac{1}{9c}}\)

\(=2+\frac{8}{3}\cdot\frac{1}{\sqrt[3]{abc}}\ge2+\frac{8}{3}\cdot\frac{1}{\frac{a+b+c}{3}}\ge10\)

Mù mắt với AM-GM cho 10 số:v

\(S=\left(a+b+c\right)+9\left(\frac{1}{9a}+\frac{1}{9b}+\frac{1}{9c}\right)\)\(\ge10\sqrt[10]{\left(a+b+c\right)\left(\frac{1}{9a}+\frac{1}{9b}+\frac{1}{9c}\right)^9}\)\(\ge10\sqrt[10]{\left(3\sqrt[3]{abc}\right)\left[3\sqrt[3]{\frac{1}{9^3abc}}\right]^9}=10\sqrt[10]{\left(3\sqrt[3]{abc}\right).\left[3^9\left(\frac{1}{9^3abc}\right)^3\right]}\)

\(=10\sqrt[10]{3^{10}.\frac{\sqrt[3]{abc}}{\left(3^6abc\right)^3}}=10\sqrt[10]{\frac{1}{3^8\sqrt[3]{\left(abc\right)^8}}}\ge10\sqrt[10]{\frac{1}{3^8\sqrt[3]{\left[\frac{\left(a+b+c\right)^3}{27}\right]^8}}}\ge10\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)

Vậy.....

Troll người tí thôi chứ bài Cauchy-Schwarz dạng Engel thì khá đơn giản:v

\(S\ge a+b+c+\frac{9}{a+b+c}=\left(a+b+c+\frac{1}{a+b+c}\right)+\frac{8}{a+b+c}\)

\(\ge2\sqrt{\left(a+b+c\right).\frac{1}{\left(a+b+c\right)}}+\frac{8}{1}=2+8=10\)

Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)

Từ giả thiết và BĐT AM-GM suy ra:\(\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)\(\ge\)3

Ta có:

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

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

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

\(\ge\)\(\frac{2}{3}\)(a+b+c-\(\frac{a}{2}\)-\(\frac{b}{2}\)-\(\frac{c}{2}\))

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

\(\ge\)\(\frac{3\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}{3}\)-1\(\ge\)2

Vậy:MinP=2 khi a=b=c=2

cách này dễ hiểu hơn nè :

Áp dụng BĐT : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)

Ta có : \(1\ge\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\ge\frac{9}{a+b+c+3}\)

\(\Leftrightarrow1\ge\frac{9}{a+b+c+3}\)\(\Leftrightarrow a+b+c+3\ge9\)\(\Leftrightarrow a+b+c\ge6\)

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

Tương tự : \(\frac{b^3}{b^2+bc+c^2}\ge b-\frac{b+c}{3}\); \(\frac{c^3}{c^2+ac+a^2}\ge c-\frac{a+c}{3}\)

Cộng cả 3 vế , ta được : \(P\ge a+b+c-\frac{2\left(a+b+c\right)}{3}=\frac{1}{3}\left(a+b+c\right)\ge\frac{1}{3}.6=2\)

Vậy GTNN của P là 2 \(\Leftrightarrow a=b=c=2\)

cảm ơn 2 bạn nhìu

ta có :

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

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

mà : \(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{a+b}\ge\frac{9}{1-a+1-b+a+b}=\frac{9}{2}\)

Vậy \(A\ge\frac{9}{2}-2=\frac{5}{2}\)

dấu bằng xảy ra khi : \(1-a=1-b=a+b\Leftrightarrow a=b=\frac{1}{3}\)

\(Q=\frac{1}{a^2+b^2}+2012+\frac{1}{ab}+4ab.\)

Ta có \(M=\frac{1}{a^2+b^2}+\frac{1}{ab}+4ab=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{1}{2ab}+8ab-4ab\)

Áp dụng bđt Cauchy ta có

\(M\ge\frac{4}{\left(a+b\right)^2}+2\sqrt{\frac{1}{2ab}.8ab}-\left(a+b\right)^2=7\)

=> \(Q\ge2012+7=2019\)

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

Vậy......

\(Q=\frac{1}{a^2+b^2}+\frac{2012ab+1}{ab}+4ab=\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\left(4ab+\frac{1}{4ab}\right)+\frac{1}{4ab}+2012\)

Áp dụng bđt \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y};\left(x+y\right)^2\ge4xy\),ta có:

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

\(\left(4ab+\frac{1}{4ab}\right)^2\ge4.4ab\cdot\frac{1}{4ab}=4\Rightarrow4ab+\frac{1}{4ab}\ge2\)

\(\left(a+b\right)^2\ge4ab\Rightarrow\frac{1}{ab}\ge\frac{4}{\left(a+b\right)^2}\ge\frac{4}{1}=4\Rightarrow\frac{1}{4ab}\ge1\)

\(\Rightarrow Q\ge4+2+1+2012=2019\)

Dấu "=" xảy ra khi a=b=1/2

Ta có : (a-b)^2 >= 0 với mọi a,b

<=> a^2-2ab+b^2 >= 0

<=> a^2+b^2 >= 2ab

<=> a^2+2ab+b^2 >= 4ab

<=> (a+b)^2 >= 4ab

Với a,b > 0 thì ta chia 2 vế cho ab .(+b) được :

a+b/ab >= 4/a+b

<=>1/a + 1/b >=4ab

Áp dụng bđt trên thì A >= 4/(a^2+b^2+2ab) = 4/(a+b)^2 >= 4/1^2 = 4

Dấu "=" xảy ra <=> a=b ; a+b =1  <=> a=b=1/2

Vậy Min A = 4 <=> x = y= 1/2

`a+ble1<=>(a+b)^2le1`

Áp dụng bđt `1/(a)+1/bge4/(a+b)` ta có:

`Age4/(a^2+2ab+b^2)=4/(a+b)^2=4/1=4`

Dấu `=` xảy ra khi:`a^2+b^2=2ab<=>(a-b)^2=0<=>a=b` và `a+b=1`

`<=>a=b=1/2`

Vậy GTNN của `A=4` khi và chỉ khi `a=b=1/2`