CMR nếu a+b+c=1 và a.b.c>0 thì ( \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \)) >= 9
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
17 tháng 11 2021
\(1,a+b+c=0\Leftrightarrow a=-b-c\Leftrightarrow a^2=b^2+2bc+c^2\Leftrightarrow b^2+c^2=a^2-2bc\)
Tương tự: \(\left\{{}\begin{matrix}a^2+b^2=c^2-2ab\\c^2+a^2=b^2-2ac\end{matrix}\right.\)
\(\Leftrightarrow N=\dfrac{a^2}{a^2-a^2+2bc}+\dfrac{b^2}{b^2-b^2+2ca}+\dfrac{c^2}{c^2-c^2+2ac}\\ \Leftrightarrow N=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ac}+\dfrac{c^2}{2bc}=\dfrac{a^3+b^3+c^3}{2abc}=\dfrac{a^3+b^3+c^3-3abc+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)+3abc}{2abc}\\ \Leftrightarrow N=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 1
Ta có:
\(\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\left(\dfrac{c}{a-b}+\dfrac{a}{b-c}+\dfrac{b}{c-a}\right)\)
\(=\dfrac{c}{a-b}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)+\dfrac{a}{b-c}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)+\dfrac{b}{c-a}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\)
Xét:
\(\dfrac{c}{a-b}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)\)
\(=1+\dfrac{c}{a-b}\left[\dfrac{b\left(b-c\right)+a\left(c-a\right)}{ab}\right]=1+\dfrac{c}{a-b}\left(\dfrac{b^2-bc+ac-a^2}{ab}\right)\)
\(=1+\dfrac{c}{a-b}\left[\dfrac{\left(b-a\right)\left(b+a\right)-c\left(b-a\right)}{ab}\right]=1+\dfrac{c}{a-b}.\dfrac{\left(b-a\right)\left(a+b-c\right)}{ab}\)
\(=1-\dfrac{c\left(a+b-c\right)}{ab}=1-\dfrac{c.\left(-2c\right)}{ab}=1+\dfrac{2c^2}{ab}\) (do \(a+b+c=0\Rightarrow a+b=-c\))
Tương tự:
\(\dfrac{a}{b-c}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{2a^2}{bc}\)
\(\dfrac{b}{c-a}\left(\dfrac{a-b}{c}+\dfrac{b-c}{a}+\dfrac{c-a}{b}\right)=1+\dfrac{2b^2}{ca}\)
\(\Rightarrow P=3+2\left(\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\right)=3+\dfrac{2\left(a^3+b^3+c^3\right)}{abc}\)
Mặt khác ta có đằng thức quen thuộc:
Khi \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)
\(\Rightarrow P=3+\dfrac{2.3abc}{abc}=9\)
27 tháng 12 2020
\(\left(a^2-bc\right)\left(b-abc\right)=\left(b^2-ca\right)\left(a-abc\right)\)
\(\Leftrightarrow a^2b+ab^2c^2-a^3bc-b^2c=b^2a+a^2bc^2-ca^2-ab^3c\)
\(\Leftrightarrow a^2b-ab^2-b^2c+ca^2=a^2bc^2-ab^3c+a^3bc-ab^2c^2\)
\(\Leftrightarrow\left(a-b\right)\left(ab+bc+ca\right)=abc\left(a-b\right)\left(a+b+c\right)\)
\(\Leftrightarrow ab+bc+ca=abc\left(a+b+c\right)\Leftrightarrow a+b+c=\dfrac{ab+bc+ca}{abc}=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(đpcm\right)\)
30 tháng 12 2020
2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)
30 tháng 12 2020
1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).
CM:....
Đặt 2x = x', 2z = z'.
Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)
\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)
\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)
\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)
\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)
PA
9 tháng 3 2017
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\) (ĐKXĐ: \(a\ne0;b\ne0;c\ne0;a+b+c\ne0\))
<=> \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)
<=> \(\dfrac{b}{ab}+\dfrac{a}{ab}+\dfrac{a+b+c}{c\left(a+b+c\right)}-\dfrac{c}{c\left(a+b+c\right)}=0\)
<=> \(\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)
<=> \(\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right)=0\)
<=> \(\left(a+b\right)\left[\dfrac{c\left(a+b+c\right)}{abc\left(a+b+c\right)}+\dfrac{ab}{abc\left(a+b+c\right)}\right]=0\)
<=> \(\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc\left(a+b+c\right)}=0\) [vì c(a + b + c) + ab = ac + bc + c2 + ab = a(b + c) + c(b + c) = (a + c)(b + c)]
<=> (a + b)(b + c)(a + c) = 0
TN
9 tháng 3 2017
câu c : vì nhân hai vế ta được :
(a+b+c)x (ab+bc+ac)=abc
abc+a\(^2\)b+\(a^2\)c + b^2c+ab^2+abc+bc^2+ac^c+abc=abc
abc+a^2b+a^2c+ b^2c+ab^2+abc+bc^2+ac^c=0
(a+c)(a+b)(b+c)=0
Lớp 9 chưa học cauchy thì làm cách này nha :v
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{a+b+c}{a}+\dfrac{a+b+c}{b}+\dfrac{a+b+c}{c}\)
\(=1+\dfrac{b}{a}+\dfrac{c}{a}+1+\dfrac{a}{b}+\dfrac{c}{b}+1+\dfrac{a}{c}+\dfrac{b}{c}\)
\(=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)+\left(\dfrac{c}{a}+\dfrac{a}{c}\right)\ge3+2+2+2=9\)
\(-->đpcm\) \("="\) khi \(a=b=c=\dfrac{1}{3}\)
áp dụng cauchy-schwarz dạng engel ta có :
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{9}{a+b+c}=9\left(đpcm\right)\)