Ta có: \(4a^2-4b^2-\frac{3a}{b}-\frac{2b}{a}+8a+12b=4\left(a^2-b^2\right)-\frac{a}{b}-2\left(\frac{a}{b}+\frac{b}{a}\right)+8a+12b\)

\(=4\left(a-b\right)\left(a+b\right)-\frac{a}{b}-2\left(\frac{a}{b}+\frac{b}{a}\right)+8a+12b\)

\(\le4\left(a-b\right)-\frac{a}{b}-2\times2+8a+12b\)

\(\le4\left(a-b\right)-\frac{a\left(a+b\right)}{b}+8a+12b-4\)

\(=4a-4b-\frac{a^2}{b}-a+8a+12b-4\)

\(=11a+9b-4-\left(\frac{a^2}{b}+b\right)\)

\(\le11a+9a-4-2a\)

\(=9\left(a+b\right)-4\)

\(=5\)

Dấu "=" xảy ra khi a = b = 0,5

help me

cíuuuuuuTvT

ta có 

\(y=2x+\frac{1}{x^2}-2\)

hay \(y=x+x+\frac{1}{x^2}-2\ge3\sqrt[3]{\frac{x.x.1}{x^2}}-2=3-2=1\)

vậy giá trị nhỏ nhất của y là 1 

Dấu bằng xảy ra khi \(x=\frac{1}{x^2}\Leftrightarrow x=1\)

ta có

\(4a^2-4b^2-\frac{3a}{b}-\frac{2b}{a}+8a+12b=4\left(a-b\right)\left(a+b\right)-\frac{a\left(a+b\right)}{b}-2\left(\frac{a}{b}+\frac{b}{a}\right)+8a+12b\)

\(\le4a-4b-\frac{a^2}{b}-2.2+7a+12b=-\frac{a^2}{b}-b+11a+9b-4\le-2a+11a+9b-4\le9-4=5\)dấu bằng xảy ra khi \(a=b=\frac{1}{2}\)

giả sử ta có hình bình hành ABCD 

ta có \(\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}\Rightarrow AC^2=AB^2+BC^2+2.AB.BC.cos\left(BAD\right)\)

\(\overrightarrow{BD}=\overrightarrow{BC}+\overrightarrow{CD}\Rightarrow BD^2=BC^2+CD^2+2BC.CDcos\left(ABC\right)\)

Nên \(AC^2+BD^2=AB^2+BC^2+CD^2+AD^2+2AB.BC.\left[cos\left(ABC\right)+cos\left(BAD\right)\right]\)

\(=AB^2+BC^2+CD^2+AD^2\)

do đó ta có điều phải chứng minh

ta có \(\hept{\begin{cases}\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}\Rightarrow AC^2=AB^2+BC^2+2\overrightarrow{AB}.\overrightarrow{BC}\\\overrightarrow{BD}=\overrightarrow{BA}+\overrightarrow{AD}\Rightarrow BD^2=BA^2+AD^2+2\overrightarrow{BA}.\overrightarrow{AD}\end{cases}}\)

mà \(\overrightarrow{AB}.\overrightarrow{BC}+\overrightarrow{BA}.\overrightarrow{AD}=\overrightarrow{AB}.\overrightarrow{BC}+\overrightarrow{AB}.\overrightarrow{AD}=0\)

Do đó \(AC^2+BD^2=2AB^2+2BC^2\Leftrightarrow m^2+n^2=2\left(a^2+b^2\right)\)