Question

[Cao long]

cho a, b ,c la 3 canh cua tam giac

chung minh (a+b+c).(1/a+1/b+1/c)>6 

Answer 1

[Nguyễn Minh Quang]

ta có \(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}\ge\frac{\left(x+y+z\right)^2}{a+b+c}.\)

áp dụng vào bài ta có\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\Rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9>6\)

Answer 2

[Xyz OLM]

Ta có :\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

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

\(=1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1\)

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

Nhận thấy \(\frac{a}{b}+\frac{b}{a}\ge2\)

Thật vậy ta có : \(\frac{a}{b}+\frac{b}{a}\ge2\)

<=> \(\frac{a^2+b^2}{ab}\ge2\Rightarrow a^2+b^2\ge2ab\Rightarrow a^2-2ab+b^2\ge0\Rightarrow\left(a-b\right)^2\ge0\left(\text{đúng}\right)\)

Tương tự ta chứng minh được \(\hept{\begin{cases}\frac{a}{c}+\frac{c}{a}\ge2\\\frac{b}{c}+\frac{c}{b}\ge2\end{cases}}\)

Khi đó \(3+\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\ge3+2+2+2\ge9>6\)(đpcm)