a + b + c= 1 \(\Rightarrow\)1 - a = b + c > 0

Tương tự : 1 - b > 0 ; 1 - c > 0

Mà 1 + a = 1 + ( 1 - b - c ) = ( 1- b ) + ( 1 - c ) \(\ge\)\(2\sqrt{\left(1-b\right)\left(1-c\right)}\)

Tương tự : \(1+b\ge2\sqrt{\left(1-a\right)\left(1-c\right)}\); \(1+c\ge2\sqrt{\left(1-a\right)\left(1-b\right)}\)

\(\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge8\sqrt{\left(1-a\right)^2\left(1-b\right)^2\left(1-c\right)^2}=8\left(1-a\right)\left(1-b\right)\left(1-c\right)\)

\(\Rightarrow A=\frac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\ge8\)

Dấu " = : xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

Vậy GTNN của A là 8 \(\Leftrightarrow a=b=c=\frac{1}{3}\)

Cách khác:

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

Áp dụng BĐT Cô si cho 2 số ta được:

\(A\ge\frac{8\sqrt{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=8\)

"=" <=> a = b = c = 1/3

Kết luận..

\(P\ge3\sqrt[3]{\dfrac{abc\left(a^2+1\right)^2\left(b^2+1\right)^2\left(c^2+1\right)^2}{a^2b^2c^2\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}}=3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)

\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(\dfrac{a+b+c}{3}\right)^3}}=9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{\left(a+b+c\right)^3}}\ge9\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{2\left(a+b+c\right)^2}}\)

Theo nguyên lý Dirichlet, trong 3 số \(a^2;b^2;c^2\) luôn có ít nhất 2 số cùng phía so với \(\dfrac{4}{9}\)

Không mất tính tổng quát, giả sử đó là \(a^2;b^2\)

\(\Rightarrow\left(a^2-\dfrac{4}{9}\right)\left(b^2-\dfrac{4}{9}\right)\ge0\)

\(\Leftrightarrow a^2b^2+\dfrac{16}{81}\ge\dfrac{4}{9}a^2+\dfrac{4}{9}b^2\)

\(\Rightarrow a^2b^2+a^2+b^2+1\ge\dfrac{13}{9}a^2+\dfrac{13}{9}b^2+\dfrac{65}{81}\)

\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\)

\(\Rightarrow\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge\dfrac{13}{9}\left(a^2+b^2+\dfrac{5}{9}\right)\left(c^2+1\right)\)

\(=\dfrac{13}{9}\left(a^2+b^2+\dfrac{4}{9}+\dfrac{1}{9}\right)\left(\dfrac{4}{9}+\dfrac{4}{9}+c^2+\dfrac{1}{9}\right)\)

\(\ge\dfrac{13}{9}\left(\dfrac{2}{3}a+\dfrac{2}{3}b+\dfrac{2}{3}c+\dfrac{1}{9}\right)^2\)

\(\Rightarrow P\ge9\sqrt[3]{\dfrac{\dfrac{13}{9}\left(\dfrac{2}{3}\left(a+b+c\right)+\dfrac{1}{9}\right)^2}{2\left(a+b+c\right)^2}}=9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9\left(a+b+c\right)}\right)^2}\)

\(P\ge9\sqrt[3]{\dfrac{13}{18}\left(\dfrac{2}{3}+\dfrac{1}{9.2}\right)^2}=\dfrac{13}{2}\)

\(P_{min}=\dfrac{13}{2}\) khi \(a=b=c=\dfrac{2}{3}\)

Từ giả thiết \(2\ge a+b+c\ge3\sqrt[3]{abc}\Rightarrow\sqrt[3]{abc}\le\dfrac{2}{3}\)

\(P\ge3\sqrt[3]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}}\)

Đặt \(Q=\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{abc}\)

\(=\dfrac{a^2b^2c^2+\left(a^2b^2+b^2c^2+c^2a^2\right)+\left(a^2+b^2+c^2\right)+1}{abc}\)

\(\ge\dfrac{a^2b^2c^2+3\sqrt[3]{\left(a^2b^2c^2\right)^2}+3\sqrt[3]{a^2b^2c^2}+1}{abc}=\dfrac{\left(\sqrt[3]{a^2b^2c^2}+1\right)^3}{abc}\)

\(=\left(\dfrac{\sqrt[3]{a^2b^2c^2}}{\sqrt[3]{abc}}+\dfrac{1}{\sqrt[3]{abc}}\right)^3=\left(\sqrt[3]{abc}+\dfrac{1}{\sqrt[3]{abc}}\right)^3\)

\(=\left(\sqrt[3]{abc}+\dfrac{4}{9\sqrt[3]{abc}}+\dfrac{5}{9\sqrt[3]{abc}}\right)^3\ge\left(2\sqrt[]{\dfrac{4\sqrt[3]{abc}}{9\sqrt[3]{abc}}}+\dfrac{5}{9.\dfrac{2}{3}}\right)^3=\dfrac{2197}{216}\)

\(\Rightarrow P\ge3\sqrt[3]{\dfrac{2197}{216}}=\dfrac{13}{2}\)

Có: \(1=\left(a+b\right)^2\le\left(a^2+b^2\right)\left(1+1\right)=2\left(a^2+b^2\right)\)

Theo bđt Bunhiacopxki có: \(\left(\text{ax}+by\right)\le\left(a^2+b^2\right)\left(x^2+y^2\right)\)

Dấu '=' xảy ra khi ay=bx

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

Dấu '=' xảy ra khi a=b=1/2

Khi đó : \(P=1:\frac{1}{4}+40.\frac{1}{8}=9\)

một cách khác :))

Áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

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

Tiếp tục áp dụng bất đẳng thức Cauchy-Schwarz dạng Engel ta có :

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

Từ (1) và (2) => \(a^4+b^4\ge\frac{\left(a^2+b^2\right)^2}{2}\ge\frac{\left(\frac{1}{2}\right)^2}{2}=\frac{1}{8}\)(3)

Theo bất đẳng thức AM-GM ta có \(ab\le\left(\frac{a+b}{2}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)=> \(\frac{1}{ab}\ge4\)(4)

Từ (3) và (4) => \(P=\frac{1}{ab}\cdot40\left(a^4+b^4\right)\ge4\cdot40\cdot\frac{1}{8}=20\)

Đẳng thức xảy ra <=> a = b = 1/2

Vậy MinP = 20

Cách khác mà kết quả khác vậy, vậy cái nào mới đúng?

\(Q=\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+bc}}\ge\sum\dfrac{\left(a+b\right)^2}{\sqrt{2\left(b+c\right)^2+\dfrac{1}{4}\left(b+c\right)^2}}=\dfrac{2}{3}\sum\dfrac{\left(a+b\right)^2}{b+c}\)

\(Q\ge\dfrac{2}{3}.\dfrac{\left(a+b+b+c+c+a\right)^2}{a+b+b+c+c+a}=\dfrac{4}{3}\left(a+b+c\right)=\dfrac{4}{3}\)

\(A=\frac{1}{a^2\left(b+c\right)}+\frac{1}{b^2\left(c+a\right)}+\frac{1}{c^2\left(a+b\right)}\)

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

\(=\frac{bc}{ab+ac}+\frac{ac}{bc+ba}+\frac{ab}{ac+bc}\)

Đặt: \(ab=x;bc=y;ac=z\)=> xyz = 1; x,y,z>0

\(A=\frac{y}{x+z}+\frac{z}{y+x}+\frac{x}{z+y}=\frac{y^2}{xy+yz}+\frac{z^2}{yz+xz}+\frac{x^2}{zx+xy}\)

\(\ge\frac{\left(x+y+z\right)^2}{2\left(xy+xz+xz\right)}\ge\frac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}=\frac{3}{2}\)

Dấu "=" xảy ra <=> x = y = z= 1 => a = b = c = 1

Vậy gtnn của A = 3/2 tại  a = b = c = 1

Ồ sorry bạn nhiều, chỗ đấy bị lỗi kĩ thuật rồi, mình sửa lại nhé :

\(M\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(ab+bc+ca\right)^2}{2\left(ab+bc+ca\right)}=\frac{ab+bc+ca}{2}\)

Lại có : \(\frac{ab+bc+ca}{2}\ge\frac{3\sqrt{a^3b^3c^3}}{2}=\frac{3}{2}\)

Do đó : \(M\ge\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

Ta có : \(\frac{1}{a^3\left(b+c\right)}=\frac{\frac{1}{a^2}}{a\left(b+c\right)}=\frac{\left(\frac{1}{a}\right)^2}{a\left(b+c\right)}\)

Tương tự : \(\frac{1}{b^3\left(a+c\right)}=\frac{\left(\frac{1}{b}\right)^2}{b\left(a+c\right)}\) , \(\frac{1}{c^3\left(a+b\right)}=\frac{\left(\frac{1}{c}\right)^2}{c\left(a+b\right)}\)

Ta thấy : \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

\(\Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)

\(\Leftrightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

Áp dụng BĐT Svacxo ta có :

\(M=\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^2\left(a+c\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)   \(\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=1\)

Vâỵ \(M_{min}=\frac{3}{2}\) tại \(a=b=c=1\)

giải thích cho mình với, sao \(\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(ab+bc+ca\right)}=\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ca\right)}\)  vậy