\(\frac{a}{a'}\)+\(\frac{b'}{b}\)=1;\(\frac{b}{b'}\)+\(\frac{c'}{c}\)=1 CMR : abc+a'b'c'=0
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26 tháng 1 2017
a) \(A=\left(1+\frac{b^2+c^2-a^2}{2bc}\right).\frac{1+\frac{a}{b+c}}{1-\frac{a}{b+c}}.\frac{b^2+c^2-\left(b-c\right)^2}{a+b+c}\)
\(=\frac{2bc+b^2+c^2-a^2}{2bc}.\frac{\frac{a+b+c}{b+c}}{\frac{b+c-a}{b+c}}.\frac{b^2+c^2-b^2+2bc-c^2}{a+b+c}\)
\(=\frac{\left(b+c+a\right)\left(b+c-a\right)}{2bc}.\frac{a+b+c}{b+c-a}.\frac{2bc}{a+b+c}\)
\(=a+b+c\)
b) \(B=\frac{\frac{3a}{a+b}}{\frac{2a}{a^2-2ab+b^2}}\)\(=\frac{3a}{a+b}.\frac{\left(a-b\right)^2}{2a}=\frac{3\left(a-b\right)^2}{2\left(a+b\right)}\)
c) \(C=\frac{\frac{a}{b}+\frac{b}{a}}{\frac{a}{b}-\frac{b}{a}}:\frac{\frac{a^2}{b^2}-\frac{b^2}{a^2}}{\left(\frac{1}{a}+\frac{1}{b}\right)^2}\)
\(=\frac{\frac{a^2+b^2}{ab}}{\frac{a^2-b^2}{ab}}:\frac{\frac{a^4-b^4}{a^2b^2}}{\frac{\left(a+b\right)^2}{a^2b^2}}\)
\(=\frac{a^2+b^2}{a^2-b^2}.\frac{\left(a+b\right)^2}{a^4-b^4}\)
\(=\frac{\left(a^2+b^2\right)\left(a+b\right)^2}{\left(a+b\right)\left(a-b\right)\left(a^2+b^2\right)\left(a+b\right)\left(a-b\right)}\)
\(=\frac{1}{\left(a-b\right)^2}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 5 2021
Ta chứng minh BĐT sau với các số dương:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Áp dụng:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế với vế:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
NV
Nguyễn Việt Lâm
Giáo viên
6 tháng 5 2021
b.
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)
Cộng vế với vế (1); (2) và (3):
\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
10 tháng 5 2016
\(D=\left(\frac{a-b}{a^{\frac{3}{4}}+a^{\frac{1}{2}}.b^{\frac{1}{4}}}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right):\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{a}{b}}\)
\(=\left[\frac{a-b}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}-\frac{a^{\frac{1}{2}}-b^{\frac{1}{2}}}{a^{\frac{1}{4}}+b^{\frac{1}{4}}}\right]:\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)^{-1}\sqrt{\frac{b}{a}}\)
\(=\frac{a-b-a+a^{\frac{1}{2}}.b^{\frac{1}{2}}}{a^{\frac{1}{2}}\left(a^{\frac{1}{4}}+b^{\frac{1}{4}}\right)}.\frac{1}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}=\frac{b^{\frac{1}{2}}}{a^{\frac{1}{2}}}\frac{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}{\left(a^{\frac{1}{4}}-b^{\frac{1}{4}}\right)}\sqrt{\frac{a}{b}}.\sqrt{\frac{a}{b}}=1\)
AH
Akai Haruma
Giáo viên
2 tháng 12 2019
Bài 1:
\(A=\frac{1}{a-b}+\frac{1}{a+b}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{a+b+a-b}{(a-b)(a+b)}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}=\frac{2a}{a^2-b^2}+\frac{2a}{a^2+b^2}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=(2a).\frac{a^2+b^2+a^2-b^2}{(a^2-b^2)(a^2+b^2)}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=\frac{4a^3}{a^4-b^4}+\frac{4a^3}{a^4+b^4}+\frac{8a^7}{a^8+b^8}\)
\(=4a^3.\frac{a^4+b^4+a^4-b^4}{(a^4-b^4)(a^4+b^4)}+\frac{8a^7}{a^8+b^8}=\frac{8a^7}{a^8-b^8}+\frac{8a^7}{a^8+b^8}=8a^7.\frac{a^8+b^8+a^8-b^8}{(a^8-b^8)(a^8+b^8)}\)
\(=\frac{16a^{15}}{a^{16}-b^{16}}\)
--------------
\(B=\frac{1}{a(a+1)}+\frac{1}{(a+1)(a+2)}+\frac{1}{(a+2)(a+3)}=\frac{(a+1)-a}{a(a+1)}+\frac{(a+2)-(a+1)}{(a+1)(a+2)}+\frac{(a+3)-(a+2)}{(a+2)(a+3)}\)
\(=\frac{1}{a}-\frac{1}{a+1}+\frac{1}{a+1}-\frac{1}{a+2}+\frac{1}{a+2}-\frac{1}{a+3}\)
\(=\frac{1}{a}-\frac{1}{a+3}=\frac{3}{a(a+3)}\)
AH
Akai Haruma
Giáo viên
2 tháng 12 2019
Bài 2:
Bạn tham khảo lời giải tương tự tại link sau:
Câu hỏi của Law Trafargal - Toán lớp 8 | Học trực tuyến
T
13 tháng 2 2020
a) Áp dụng BĐT Cauchy-Schwarz dạng Engel: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Tương tự:\(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c};\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)
Cộng theo vế 3 BĐT trên rồi chia cho 2 ta thu được đpcm
Đẳng thức xảy ra khi \(a=b=c\)
b)Đặt \(a+b=x;b+c=y;c+a=z\). Cần chứng minh:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
Cách làm tương tự câu a.
c) \(VT=\Sigma_{cyc}\frac{1}{\left(a+b\right)+\left(a+c\right)}\le\frac{1}{4}\Sigma_{cyc}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\le\frac{1}{16}\Sigma\left(\frac{2}{a}+\frac{1}{b}+\frac{1}{c}\right)=1\)
Đẳng thức xảy ra khi \(a=b=c=\frac{3}{4}\)
d) Em làm biếng quá anh làm nốt đi:P
9 tháng 11 2016
a/ \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ; \(\frac{1}{b}+\frac{1}{c}\ge\frac{4}{b+c}\) ; \(\frac{1}{c}+\frac{1}{a}\ge\frac{4}{c+a}\)
Cộng theo vế :
\(2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)
b/ \(\frac{1}{a+b}+\frac{1}{b+c}\ge\frac{4}{a+2b+c}\)
\(\frac{1}{b+c}+\frac{1}{c+a}\ge\frac{4}{b+2c+a}\)
\(\frac{1}{c+a}+\frac{1}{a+b}\ge\frac{4}{c+b+2a}\)
Cộng theo vế :
\(2\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\ge4\left(\frac{1}{2a+b+c}+\frac{1}{2b+c+a}+\frac{1}{2c+a+b}\right)\)
\(\Leftrightarrow\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\ge2\left(\frac{1}{2a+b+c}+\frac{1}{a+2b+c}+\frac{1}{a+b+2c}\right)\)
Q
10 tháng 7 2017
\(\dfrac{1}{\left(a+b\right)^3}\left(\dfrac{1}{a^3}+\dfrac{1}{a^3}\right)+\dfrac{3}{\left(a+b\right)^4}+\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)+\dfrac{6}{\left(a+b\right)^5}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)
\(=\dfrac{1}{\left(a+b\right)^3}\cdot\dfrac{b^3+a^3}{a^3b^3}+\dfrac{3}{\left(a+b\right)^4}\cdot\dfrac{b^2+a^2}{a^2b^2}+\dfrac{6}{\left(a+b\right)^5}\cdot\dfrac{b+a}{ab}\)
\(=\dfrac{1}{\left(a+b\right)^3}\cdot\dfrac{\left(b+a\right)\left(a^2-ab+a^2\right)}{a^3b^3}+\dfrac{3\left(b^2+a^2\right)}{a^2b^2\cdot\left(a+b\right)^4}\cdot\dfrac{6}{\left(a+b\right)^4}\cdot\dfrac{1}{ab}\)
\(=\dfrac{1}{\left(a+b\right)^2}\cdot\dfrac{b^2-ab+a^2}{a^3b^3}+\dfrac{3b^2+3a^2}{a^2b^2\cdot\left(a+b\right)^4}+\dfrac{6}{ab\left(a+b\right)^4}\)
\(=\dfrac{b^2-ab+a^2}{a^3b^3\cdot\left(a+b\right)^2}+\dfrac{3b^2+3a^2}{a^2b^2\cdot\left(a+b\right)^4}+\dfrac{6}{ab\cdot\left(a+b\right)^4}\)
\(=\dfrac{\left(a+b\right)^2\cdot\left(b^2-ab+a^2\right)+ab\left(3b^2+3a^2\right)+6a^2b^2}{a^3b^3\cdot\left(a+b\right)^4}\)
\(=\dfrac{\left(a^2+2ab+b^2\right)\left(b^2-ab+a^2\right)+3ab^3+3a^3b+6a^2b^2}{a^3b^3\cdot\left(a+b\right)^4}\)
\(=\dfrac{a^2b^2-a^3b+a^4+2ab^3-2a^2b^2+2a^3b+b^4-ab^3+a^2b^2+3ab^3+3a^2b+6a^2b^2}{a^3b^3\cdot\left(a+b\right)^4}\)
\(=\dfrac{6a^2b^2+4a^3b+a^4+4ab^3+b^4}{a^3b^3\cdot\left(a+b\right)^4}\)
Ta có:
\(\frac{a}{a'}+\frac{b'}{b}=1\)
\(\Rightarrow\frac{a}{a'}.\frac{b}{b'}+\frac{b'}{b}.\frac{b}{b'}=\frac{b}{b'}\)
\(\Rightarrow\frac{ab}{a'b'}+\frac{b'b}{bb'}=\frac{b}{b'}.\)
\(\Rightarrow\frac{ab}{a'b'}+1=\frac{b}{b'}\) (1).
Lại có:
\(\frac{b}{b'}+\frac{c'}{c}=1\)
\(\Rightarrow\frac{b}{b'}=1-\frac{c'}{c}\) (2).
Từ (1) và (2) \(\Rightarrow\frac{ab}{a'b'}+1=1-\frac{c'}{c}.\)
\(\Rightarrow\frac{ab}{a'b'}=-\frac{c'}{c}.\)
\(\Rightarrow abc=-a'b'c'\)
\(\Rightarrow abc+a'b'c'=0\left(đpcm\right).\)
Chúc bạn học tốt!