Toán Cho a,b,c>0.CM:(a+1/b).(b+1/c).(c+1/a)>=8 26/07/2021 By Jade Cho a,b,c>0.CM:(a+1/b).(b+1/c).(c+1/a)>=8
Ta có $a + \dfrac{1}{b} \geq 2\sqrt{\dfrac{a}{b}}$ (Cauchy) CMTT ta cũng có $b + \dfrac{1}{c} \geq 2\sqrt{\dfrac{b}{c}}$ $c + \dfrac{1}{a} \geq 2\sqrt{\dfrac{c}{a}}$ Vậy ta có $\left( a + \dfrac{1}{b} \right) \left( b + \dfrac{1}{c} \right) \left( c + \dfrac{1}{a} \right) \geq 2^3 . \sqrt{\dfrac{a}{b} . \dfrac{b}{c} . \dfrac{c}{a}} = 8$ Dấu “=” xảy ra khi $a = \dfrac{1}{b}, b = \dfrac{1}{c}, c = \dfrac{1}{a}$ hay $a = b=c=1$. Trả lời
Đáp án:
Giải thích các bước giải:
Ta có
$a + \dfrac{1}{b} \geq 2\sqrt{\dfrac{a}{b}}$ (Cauchy)
CMTT ta cũng có
$b + \dfrac{1}{c} \geq 2\sqrt{\dfrac{b}{c}}$
$c + \dfrac{1}{a} \geq 2\sqrt{\dfrac{c}{a}}$
Vậy ta có
$\left( a + \dfrac{1}{b} \right) \left( b + \dfrac{1}{c} \right) \left( c + \dfrac{1}{a} \right) \geq 2^3 . \sqrt{\dfrac{a}{b} . \dfrac{b}{c} . \dfrac{c}{a}} = 8$
Dấu “=” xảy ra khi $a = \dfrac{1}{b}, b = \dfrac{1}{c}, c = \dfrac{1}{a}$ hay $a = b=c=1$.