\(A=\dfrac{\left(a+b+c+a\right)\left(a+b+c+b\right)\left(a+b+c+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

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

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

\(\frac{1}{a^4\left(1+b\right)\left(1+c\right)}=\frac{1}{\frac{a^4\left(1+b\right)\left(1+c\right)}{abc}}=\frac{\frac{1}{a^3}}{\left(\frac{1}{b}+1\right)\left(\frac{1}{c}+1\right)}\)

Đặt \(\left(x;y;z\right)=\left(\frac{1}{a};\frac{1}{b};\frac{1}{c}\right)\), tương tự suy ra:

\(A=\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+x\right)\left(1+z\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}\)

Theo BĐT AM-GM ta có: \(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge\frac{3x}{4}\)

Tương tự suy ra \(A+\frac{3}{4}+\frac{x+y+z}{4}\ge\frac{3\left(x+y+z\right)}{4}\)

\(\Rightarrow A\ge\frac{x+y+z}{2}-\frac{3}{4}\ge\frac{3\sqrt[3]{xyz}}{2}-\frac{3}{4}=\frac{3}{4}\)

Dấu = xảy ra khi x=y=z=1 hay a=b=c=1

VỚi các số thực: a,b,c >0 thỏa a+b+c=1. Chứng minh rằng: \(\frac{1+a}{1-a}+\frac{1+b}{1-b}+\frac{1+c}{1-c}\le2\left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)\)

Help me

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..

\(a^2b^2c^2+\left(a+1\right)\left(1+b\right)\left(1+c\right)\ge a+b+c+ab+bc+ca+3\)

\(\Leftrightarrow\left(abc\right)^2+abc-2\ge0\Leftrightarrow\left(abc+2\right)\left(abc-1\right)\ge0\Leftrightarrow abc\ge1\)

Áp dụng BĐT Cosi ta có:

\(\frac{a^3}{\left(b+2c\right)\left(2c+3a\right)}+\frac{b+2c}{45}+\frac{2c+3a}{75}\ge3\sqrt[3]{\frac{a^3}{\left(b+2c\right)\left(2c+3b\right)}\cdot\frac{b+2c}{45}\cdot\frac{2c+3a}{75}}=\frac{a}{5}\left(1\right)\)

Tương tự ta có: \(\hept{\begin{cases}\frac{b^3}{\left(c+2a\right)\left(2a+3b\right)}+\frac{c+2a}{45}+\frac{2a+3b}{75}\ge\frac{b}{5}\left(2\right)\\\frac{c^3}{\left(a+2b\right)\left(2b+3c\right)}+\frac{a+2b}{45}+\frac{2b+3c}{75}\ge\frac{c}{5}\left(3\right)\end{cases}}\)

Từ (1)(2)(3) ta có:

\(P+\frac{2\left(a+b+c\right)}{15}\ge\frac{a+b+c}{5}\Leftrightarrow P\ge\frac{1}{15}\left(a+b+c\right)\)

Mà \(a+b+c\ge3\sqrt[3]{abc}\Rightarrow S\ge\frac{1}{5}\)

Dấu "=" xảy ra <=> a=b=c=1

CHÚC BAN HỌC GIỎI

đây\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)