CMR : ( a + b +c) (\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)) \(\ge9\)
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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 <=>\(a=b=c=\dfrac{1}{3}\)
12 tháng 5 2022
-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 \(a=b=c=\dfrac{1}{3}\)
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!
ND
23 tháng 7 2018
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
\(=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)
\(=\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+3\)
Áp dụng BĐT Cô - si cho 2 số không âm:
\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\)
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}=2\)
Suy ra:
\(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{a}{c}+\dfrac{c}{a}+\dfrac{b}{c}+\dfrac{c}{b}+3\ge2+2+2+3=9\)
Dấu "=" xảy ra khi: a = b = c
23 tháng 7 2018
Áp dụng BĐT Cauchy dạng Engel , ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ≥ \(\dfrac{9}{a+b+c}\)
⇔ \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}\left(a+b+c\right).\dfrac{9}{a+b+c}\)
⇔ \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\text{≥}9\)
\("="\text{⇔}a=b=c\)
1 tháng 1 2018
Cauchy-Schwarz: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{1}=9\)
26 tháng 6 2021
do \(a,b,c\ge1\)\(=>\left\{{}\begin{matrix}b+c\ge2\\c+a\ge2\\a+b\ge2\end{matrix}\right.\)
\(=>\left\{{}\begin{matrix}a\left(b+c\right)\ge2a\\b\left(c+a\right)\ge2b\\c\left(a+b\right)\ge2c\end{matrix}\right.\)
\(=>\) biểu thức đề bài cho\(\ge2\left(a+b+c+\dfrac{1}{a^2+1}+\dfrac{1}{b^2+1}+\dfrac{1}{c^2+1}\right)\)
\(2\left(1+1+1+\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}\right)=9\)
dấu= xảy ra<=>a=b=c=1
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:
\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\)
\(\dfrac{b}{c}+\dfrac{c}{b}\ge2\)
\(\Rightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge2+2+2+3\)
\(\Rightarrow3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\ge9\)
\(\Rightarrow\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(đpcm\right)\)
Học mỗi cái \(a+b\ge2\sqrt{ab}\) này thôi hả. Không sao a chiều được
\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1+1+1+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)
\(\ge3+2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}+2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}+2\sqrt{\dfrac{b}{c}.\dfrac{c}{b}}\)
\(=3+2+2+2=9\)
Xong.
C-S kind ENgel \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{\left(1+1+1\right)^2}{a+b+c}=\dfrac{9}{a+b+c}\Rightarrow DPCM\)