Đầu tiên ta để ý \(\left(1+a\right)\left(1+b\right)\left(1+c\right)=1+\left(a+b+c\right)+\left(ab+bc+ca\right)+abc\)

\(\ge1+3\sqrt[3]{abc}+3\sqrt[3]{\left(abc\right)^2}+abc=\left(1+\sqrt[3]{abc}\right)^3.\)

Ta sử dụng bất đẳng thức quen thuộc \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0.\)

Khi đó

\(\left(\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\right)^2\ge3\times\frac{a+b+c+ab+bc+ca}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}=\frac{3}{abc}-\frac{3}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}-\frac{3}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\)

\(\ge\frac{3}{t^3}-\frac{3}{\left(t+1\right)^3}-\frac{3}{t^3\left(t+1\right)^3}=\frac{9}{t^2\left(1+t\right)^2}\)   trong đó \(t=\sqrt[3]{abc}\).

Vậy ta có \(\frac{1}{a\left(b+1\right)}+\frac{1}{b\left(c+1\right)}+\frac{1}{c\left(a+1\right)}\ge\frac{3}{t\left(t+1\right)}\ge\frac{3}{\left(t^2-t+1\right)\left(t+1\right)}=\frac{3}{t^3+1}=\frac{3}{abc+1}.\)   (ĐPCM)

thế kỉ 19 đúng

chịu thôi

a/ Chia làm 2 trường hợp :

+) x - 1 = 2x => -x = 1 => x = -1

+) x - 1 = -2x => 3x = 1 => x = 1/3 

Vậy x = -1 ; x = 1/3

b/ \(\Rightarrow x=x-5+\left(x+5\right)\left(1-x\right)\)

\(\Rightarrow x=x-5+x-x^2+5-5x\)

\(\Rightarrow x^2+4x=0\Rightarrow x\left(x+4\right)=0\)

\(\Rightarrow x=0\) hoặc \(x+4=0\Rightarrow x=-4\)

Vậy x = 0 ; x = -4