Ta có: \(abc=1\Leftrightarrow\hept{\begin{cases}ab=\frac{1}{c}\\bc=\frac{1}{a}\\ca=\frac{1}{b}\end{cases}}\)

\(abc=1\Leftrightarrow\sqrt[3]{abc}=1\)

Áp dụng BĐT AM-GM ta có:\(1=\sqrt[3]{abc}\le\frac{a+b+c}{3}\Leftrightarrow a+b+c\ge3\)

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

\(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge4\left(a+b+c-1\right)\)

\(\Leftrightarrow\)\(a^2b+ab^2+a^2c+ac^2+b^2c+cb^2+2abc+4\ge4\left(a+b+c\right)\)

\(\Leftrightarrow\frac{a}{c}+\frac{b}{c}+\frac{a}{b}+\frac{c}{b}+\frac{b}{a}+\frac{c}{a}+6\ge4\left(a+b+c\right)\)

\(\Leftrightarrow\frac{a+b}{c}+\frac{a+c}{b}+\frac{b+c}{a}+6\ge4\left(a+b+c\right)\)

\(\Leftrightarrow\frac{a+b+c}{c}+\frac{a+c+b}{b}+\frac{a+b+c}{a}+3\ge4\left(a+b+c\right)\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+3\ge4\left(a+b+c\right)\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c}\ge4\)(1)

Ta chứng mĩnh BĐT phụ

Với a,b,c > thì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)

Thật vậy.

Áp dụng BĐT AM-GM ta có:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{3}{\sqrt[3]{abc}}\ge\frac{3}{\frac{a+b+c}{3}}=\frac{9}{a+b+c}\)

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

Áp dụng \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{3}{a+b+c}\ge\frac{9}{a+b+c}+\frac{3}{a+b+c}=\frac{12}{3}=4\)(2)

Từ (1) và  (2)

=> \(\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge4\left(a+b+c-1\right)\)

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

Bạn ơi, tại sao \(\frac{9}{a+b+c}+\frac{3}{a+b+c}=\frac{12}{3}\) được hả bạn?

\(\left(x^2+2x\right)\left(x^2+2x-2\right)=3\)

\(\Leftrightarrow x^4+4x^3+2x^2-4x=3\)

\(\Leftrightarrow x^4+4x^3+2x^2-4x-3=3-3\)

\(\Leftrightarrow x^4+4x^3+2x^2-4x-3=0\)

\(\Leftrightarrow\left(x+1\right)^2\left(x+3\right)\left(x-1\right)=0\)

Dễ rồi, tự làm nốt đi

cảm ơn bạn nhé

bạn là fan của Cris phải ko

Áp dụng BĐT Cauchy-Schwarz dạng Engel,ta có:

\(A=\frac{a^2}{a+1}+\frac{b^2}{b+1}\ge\frac{\left(a+b\right)^2}{a+b+2}=\frac{1}{1+2}=\frac{1}{3}^{\left(đpcm\right)}\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}a+b=1\\\frac{a}{a+1}=\frac{b}{b+1}\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=1\\ab+a=ab+b\end{cases}}\Leftrightarrow a=b=\frac{1}{2}\)

Vậy ...

\(\left(x^2+9\right)\left(3-x\right)\left(2x-4\right)=0\)

Dễ thấy \(x^2+9>0\forall x\)do đó :

\(\orbr{\begin{cases}3-x=0\\2x-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

Vậy....

\((x^2+9)(3-x)(2x-4)=0\)

Dễ thấy : \(x^2+9>0\forall x\), do đó :

\(\hept{\begin{cases}3-x=0\\2x-4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\x=2\end{cases}}\)

Vậy :