cho a,b,c > 0. Chứng minh (a+b+c)(1/a + 1/b + 1/c) $\geq$ 9
2 câu trả lời
$\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9$
Áp dụng BĐT Cô - si:
$\left\{\begin{matrix}a+b+c\ge3\sqrt[3]{abc}\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}\end{matrix}\right.$
$\rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}.3\sqrt[3]{\frac{1}{abc}}$
$\rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9.\sqrt[3]{\frac{abc}{abc}}$
$\rightarrow\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9$
`-> ĐPCM`
$@Thanh$
`(a + b + c)(1/a + 1/b + 1/c)`
`= a/a + a/b + a/c + b/a + b/b + b/c + c/a + c/b + c/c`
`= 3 + (a/b + b/a) + (a/c + c/a) + (b/c + c/b)`
Áp dụng BĐT Cauchy cho `3` số dương `a; b; c`,ta có:
`a/b + b/a ≥ 2.`$\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}$ `= 2`
`a/c + c/a ≥ 2.`$\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}$ `= 2`
`b/c + c/b ≥ 2.`$\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}$ `= 2`
`⇒ 3 + (a/b + b/a) + (a/c + c/a) + (b/c + c/b) ≥ 3 + 2 + 2 + 2 = 9`
Dấu "`=`" xảy ra khi: $\begin{cases} \dfrac{a}{b}=\dfrac{b}{a}\\\dfrac{a}{c}=\dfrac{c}{a}\\\dfrac{b}{c}=\dfrac{c}{b} \end{cases}$
`⇔ a = b = c`