Cho a,b,c >0 Chứng minh: (a+b+c)x1 1 1
(_+_+_ ) >/ 9
a b c
Cho a,b,c >0 Chứng minh: (a+b+c)x1 1 1
(_+_+_ ) >/ 9
a b c
dBài toán : $(a+b+c).\bigg(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\bigg) ≥ 9 $
$\to 1+1+1+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b} ≥ 9 $
$\to \dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b} ≥ 6 $
Ta đi chứng minh $\dfrac{x}{y}+\dfrac{y}{x} ≥ 2 ∀x,y>0$
$⇔ \dfrac{(x-y)^2}{xy} ≥ 0 $ ( Đúng )
Áp dụng vào bài toán thì :
$\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b} ≥ 6 $
Dấu “=” xảy ra $⇔a=b=c$
Vậy ta có điều phải chứng minh.
Theo suy luận thì nó như thế này
$(a+b+c) (\dfrac{1}{a} + \dfrac{1}{b} +\dfrac{1}{c} )$ $\geq$ $9$
=> $\dfrac{a}{a} + \dfrac{a}{b} + \dfrac{a}{c} + \dfrac{b}{a}+\dfrac{b}{c}+\dfrac{b}{b} + \dfrac{c}{c}$
$+ \dfrac{c}{a} + \dfrac{c}{b}$ $\geq$ $9$
$=> 3 + ( \dfrac{a}{b} + \dfrac{b}{a}) + (\dfrac{b}{c} + \dfrac{c}{b}) + ( \dfrac{a}{c} + \dfrac{c}{a})$ $\geq$ $9$
Mà theo Cauchy ta có
$\dfrac{a}{b} + \dfrac{b}{a} \geq 2 \sqrt[]{\dfrac{a}{b} . \dfrac{b}{a}}$ $= 2$
Tương tự
$\dfrac{a}{c} + \dfrac{c}{a} \geq 2 \sqrt[]{\dfrac{a}{c} . \dfrac{c}{a}}$ $= 2$
$\dfrac{b}{c} + \dfrac{c}{b} \geq 2 \sqrt[]{\dfrac{b}{c} . \dfrac{c}{b}}$ $= 2$
$=> (a+b+c) (\dfrac{1}{a} + \dfrac{1}{b} +\dfrac{1}{c} )$ $\geq$ $3+2+2+2= 9$
Dấu $”=”$ khi $a = b = c$