Cho 3 số dương \(a,b,c\) và \(a+b+c=1\)
CMR \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
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NQ
4
30 tháng 4 2017
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}=3+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)Ta có: \(\dfrac{a}{b}+\dfrac{b}{a}\ge2;\dfrac{c}{a}+\dfrac{a}{c}\ge2;\dfrac{b}{c}+\dfrac{c}{b}\ge2\)
\(\Leftrightarrow\)\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3+2+2+2=9\)
NM
30 tháng 4 2017
Gọi \(A=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) Ta có:
\(A=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)\(=\left(\dfrac{a}{a}+\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+\dfrac{b}{b}+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{b}{c}+\dfrac{c}{c}\right)\)
\(=\left(1+\dfrac{b}{a}+\dfrac{c}{a}\right)+\left(\dfrac{a}{b}+1+\dfrac{c}{b}\right)+\left(\dfrac{a}{c}+\dfrac{b}{c}+1\right)\)
\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\)
\(=3+\left(\dfrac{a^2}{ab}+\dfrac{b^2}{ab}\right)+\left(\dfrac{b^2}{bc}+\dfrac{c^2}{bc}\right)+\left(\dfrac{c^2}{ac}+\dfrac{a^2}{ac}\right)\)
\(=3+\dfrac{a^2+b^2}{ab}+\dfrac{b^2+c^2}{bc}+\dfrac{c^2+a^2}{ac}\)
\(=3+\dfrac{a^2-2ab+b^2+2ab}{ab}+\dfrac{b^2-2bc+c^2+2bc}{bc}+\dfrac{c^2-2ac+a^2+2ac}{ac}\)
\(=3+\dfrac{\left(a-b\right)^2+2ab}{ab}+\dfrac{\left(b-c\right)^2+2bc}{bc}+\dfrac{\left(c-a\right)^2+2ac}{ac}\)
\(=3+\dfrac{\left(a-b\right)^2}{ab}+2+\dfrac{\left(b-c\right)^2}{bc}+2+\dfrac{\left(c-a\right)^2}{ac}+2\)
\(=9+\dfrac{\left(a-b\right)^2}{ab}+\dfrac{\left(b-c\right)^2}{bc}+\dfrac{\left(c-a\right)^2}{ac}\)
Ta thấy: \(\dfrac{\left(a-b\right)^2}{ab}\ge0\) với \(\forall\) a, b
\(\dfrac{\left(b-c\right)^2}{bc}\ge0\) với \(\forall\) b, c
\(\dfrac{\left(c-a\right)^2}{ac}\ge\) 0 với \(\forall\) a, c
=> \(A\ge9\).
Vậy...
Chúc bạn học tốt!
1 tháng 1 2018
Cauchy-Schwarz: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{1}=9\)
SW
2
DK
30 tháng 5 2021
Áp dụng AM-GM
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\sqrt[3]{abc}.3.\dfrac{1}{\sqrt[3]{abc}}=9\)
\(\rightarrow1.\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)
vậy ta có điều phải chứng minh
Dấu "=" \(a=b=c=\dfrac{1}{3}\)
30 tháng 5 2021
Áp dụng svac-xơ:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=9\)
Dấu = xảy ra <=> \(a=b=c=\dfrac{1}{3}\)
C2: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)
\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\)
\(\ge3+2+2+2=9\) (theo cosi)
Dấu = xảy ra <=>\(a=b=c=\dfrac{1}{3}\)
PM
cho 3 số dương a,b,c có tổng bằng 1
chứng minh rằng : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
2
29 tháng 1 2023
Mình bổ sung một cách làm khác nhé.
Áp dụng BĐT Cô-si cho 3 số dương \(a,b,c\), ta có \(a+b+c\ge3\sqrt[3]{abc}\) \(\Rightarrow1\ge3\sqrt[3]{abc}\) (1)
Áp dụng BĐT Cô-si cho 3 số dương \(\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\) ta có \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{\dfrac{1}{abc}}\) (2)
Nhân theo vế của các BĐT (1) và (2), ta được \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\sqrt[3]{abc}.3\sqrt[3]{\dfrac{1}{abc}}=9\) (đpcm)
Đẳng thức xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)
29 tháng 1 2023
\(Ta\) có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)
\(=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)
\(=1+\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{a}{b}+\dfrac{c}{b}+1+\dfrac{a}{c}+\dfrac{b}{c}+1\)
\(=\left(1+1+1\right)+\left(\dfrac{b}{a}+\dfrac{a}{b}\right)+\left(\dfrac{c}{b}+\dfrac{b}{c}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\)
\(Ta\) có : \(\left(\dfrac{b}{a}+\dfrac{a}{b}\right)\ge2\Leftrightarrow\dfrac{a^2+b^2}{ab}-2\ge0\Leftrightarrow\dfrac{a^2-2ab+b^2}{ab}\ge0\)
\(cmt\) \(tương\) \(tự\) \(với\) : \(\left(\dfrac{c}{b}+\dfrac{b}{c}\right)\) \(và\) \(\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\) \(đều\) \(\ge2\) \(như\) \(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge2\)
\(\Rightarrow\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\ge9\) \(hay\) \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge9\)
8 tháng 4 2018
Có \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)
\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
Áp dụng BĐT Cô-si, ta có:
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}\ge2\)
C/m tương tự, ta có:
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\)
\(\Rightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge2+2+2+3\)
\(\Rightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge9\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(đpcm\right)\)
15 tháng 8 2023
a,b,c là các số dương nên \(\left(a+b+c\right)>=3\cdot\sqrt[3]{abc}\)
\(\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}\)
Do đó: \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=3\cdot\sqrt[3]{abc}\cdot3\cdot\sqrt[3]{\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\cdot\sqrt[3]{a\cdot b\cdot c\cdot\dfrac{1}{a}\cdot\dfrac{1}{b}\cdot\dfrac{1}{c}}=9\)
AH
Akai Haruma
Giáo viên
8 tháng 9 2017
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq (1+1+1)^2\)
\(\Leftrightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\geq 9\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=\frac{1}{3}\)
-Áp dụng BĐT Caushy Schwarz ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{1}=9\)
-Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)