Ta chứng minh bất đẳng thức phụ sau \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)(*)

\(BĐT\)(*) \(< =>\frac{x+y}{xy}\ge\frac{4}{x+y}< =>\left(x+y\right)^2\ge4xy< =>\left(x-y\right)^2\ge0\)*đúng*

Sử dụng bất đẳng thức (*) ta có \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\); \(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c}\); \(\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)

Cộng theo vế 3 bất đẳng thức trên ta được \(\frac{2}{a}+\frac{2}{b}+\frac{2}{c}\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)

\(< =>2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge2\left(\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\right)\)

\(< =>\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{2}{a+b}+\frac{2}{b+c}+\frac{2}{c+a}\)(Dấu "=" xảy ra khi và chỉ khi \(a=b=c\))

Done!

Áp dụng bất đẳng thức \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có : ( bạn tự chứng minh )

\(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)*1*

\(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c}\)*2*

\(\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)*3*

Cộng *1* , *2* , *3* theo vế ta có :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{c}+\frac{1}{a}\ge\frac{4}{a+b}+\frac{4}{b+c}+\frac{4}{c+a}\)

<=> \(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

<=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

Vậy ta có điều phải chứng minh

Đẳng thức xảy ra <=> a = b = c > 0

\(P=\left(x^2+4x+1\right)^2-12\left(x+2\right)^2+2093\)

\(=\left(x^2+4x+4-3\right)^2-12\left(x+2\right)^2+2093\)

\(=\left[\left(x+2\right)^2-3\right]^2-12\left(x+2\right)^2+2093\)

\(=\left(x+2\right)^4-6\left(x+2\right)^2+9-12\left(x+2\right)^2+2093\)

\(=\left(x+2\right)^4-18\left(x+2\right)^2+2102\)

\(=\left(x+2\right)^4-18\left(x+2\right)^2+81+2021\)

\(=\left[\left(x+2\right)^4-18\left(x+2\right)^2+81\right]+2021\)

\(=\left[\left(x+2\right)^2-9\right]^2+2021\)

\(=\left[\left(x+2-3\right)\left(x+2+3\right)\right]^2+2021\)

\(=\left[\left(x-1\right)\left(x+5\right)\right]^2+2021\)

Vì \(\left[\left(x-1\right)\left(x+5\right)\right]^2\ge0\forall x\)

\(\Rightarrow\left[\left(x-1\right)\left(x+5\right)\right]^2+2021\ge2021\)\(\forall x\)

hay \(P\ge2021\)

Dấu " = " xảy ra \(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Vậy \(minP=2021\)\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)

Sai đề nha.

\(0^n+0^n=0^n\)

n={1;2;3}

6a( x - 3y ) - 8b( 3y - x )

= 6a( x - 3y ) + 8b( x - 3y )

= 2( x - 3y )( 3a + 4b )

\(6a\left(x-3y\right)-8b\left(3y-x\right)\)

\(=6a\left(x-3y\right)+8b\left(x-3y\right)\)

\(=\left(6a+8b\right)\left(x-3y\right)=2\left(3a+4b\right)\left(x-3y\right)\)