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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: cho a,b,c > 0. Chứng minh (a+b+c)(1/a + 1/b + 1/c) $\geq$ 9

Home/ Toán Lớp 9: cho a,b,c > 0. Chứng minh (a+b+c)(1/a + 1/b + 1/c) $\geq$ 9

Toán Lớp 9: cho a,b,c > 0. Chứng minh (a+b+c)(1/a + 1/b + 1/c) $\geq$ 9

Tháng Tám 28, 2022 Toán học 2 Comments 0 views

Toán Lớp 9: cho a,b,c > 0. Chứng minh (a+b+c)(1/a + 1/b + 1/c) $\geq$ 9

Comments ( 2 )

  1. tuyethutanhthu
    tuyethutanhthu
    28 Tháng Tám, 2022 at 7:08 chiều
    Reply
    (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
     
     

  2. landinhngoc97
    landinhngoc97
    28 Tháng Tám, 2022 at 7:08 chiều
    Reply
    $\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9$
    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
    $@Thanh$

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222-9+11+12:2*14+14 = ? ( )

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