Cho \(\dfrac{a}{c}=\dfrac{a-b}{b-c}\);a#0;c#0;a-b#0;b-c#0.
CM: \(\dfrac{1}{a}+\dfrac{1}{a-b}=\dfrac{1}{b-c}-\dfrac{1}{c}\)
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
19 tháng 2 2021
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{b}{a}-1=\dfrac{d}{c}-1\)
\(\Leftrightarrow\dfrac{b-a}{a}=\dfrac{d-c}{c}\)
\(\Leftrightarrow\dfrac{a-b}{a}=\dfrac{c-d}{c}\)
\(\Leftrightarrow\dfrac{a}{a-b}=\dfrac{c}{c-d}\)(đpcm)
3 tháng 1 2019
3/ Áp dụng bất đẳng thức AM-GM, ta có :
\(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}\ge2\sqrt{\dfrac{\left(ab\right)^2}{\left(bc\right)^2}}=\dfrac{2a}{c}\)
\(\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge2\sqrt{\dfrac{\left(bc\right)^2}{\left(ac\right)^2}}=\dfrac{2b}{a}\)
\(\dfrac{c^2}{a^2}+\dfrac{a^2}{b^2}\ge2\sqrt{\dfrac{\left(ac\right)^2}{\left(ab\right)^2}}=\dfrac{2c}{b}\)
Cộng 3 vế của BĐT trên ta có :
\(2\left(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\right)\ge2\left(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\right)\)
\(\Leftrightarrow\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\ge\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\left(\text{đpcm}\right)\)
AH
Akai Haruma
Giáo viên
4 tháng 1 2019
Bài 1:
Áp dụng BĐT AM-GM ta có:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{1}{2\sqrt{a^2.bc}}+\frac{1}{2\sqrt{b^2.ac}}+\frac{1}{2\sqrt{c^2.ab}}=\frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ac}}{2abc}\)
Tiếp tục áp dụng BĐT AM-GM:
\(\sqrt{bc}+\sqrt{ac}+\sqrt{ab}\leq \frac{b+c}{2}+\frac{c+a}{2}+\frac{a+b}{2}=a+b+c\)
Do đó:
\(\frac{1}{a^2+bc}+\frac{1}{b^2+ac}+\frac{1}{c^2+ab}\leq \frac{\sqrt{ab}+\sqrt{bc}+\sqrt{ca}}{2abc}\leq \frac{a+b+c}{2abc}\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 3 2022
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{a+b}{a+b+c}+\dfrac{b+c}{a+b+c}=2\) (1)
\(VP=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\sqrt{\dfrac{c}{a+b}}=\dfrac{a}{\sqrt{a\left(b+c\right)}}+\dfrac{b}{\sqrt{b\left(c+a\right)}}+\dfrac{c}{\sqrt{c\left(a+b\right)}}\)
\(VP\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=2\) (2)
(1);(2) \(\Rightarrow VT< VP\)
H
8 tháng 7 2021
Áp dụng bất đẳng thức Cosi ta có :
\(\frac{a}{b+c} + \frac{b+c}{4a} \geq 1;\frac{b}{c+a} + \frac{c+a}{4b} \geq 1;\frac{c}{a+b} + \frac{a+b}{4c} \geq 1\)
\(\frac{b}{a}+\frac{a}{b} \geq 2;\frac{c}{b}+\frac{b}{c} \geq 2;\frac{a}{c}+\frac{c}{a} \geq 2\)
\(\Rightarrow A=( \frac{a}{b+c} + \frac{b+c}{4a} +\frac{b}{c+a} + \frac{c+a}{4b} +\frac{c}{a+b} + \frac{a+b}{4c}) +\frac{3}{4}(\frac{b}{a}+\frac{a}{b}+\frac{c}{b}+\frac{b}{c} +\frac{a}{c}+\frac{c}{a} )\)
\(\geq 1+1+1+\frac{3}{4} (2+2+2)=\frac{15}{2}\)
Dấu = xảy ra khi và chỉ khi a=b=c>0
30 tháng 11 2021
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}\\ \Rightarrow\dfrac{b+c+d}{a}=\dfrac{a+c+d}{b}=\dfrac{a+b+d}{c}=\dfrac{a+b+c}{d}=\dfrac{3\left(a+b+c+d\right)}{a+b+c+d}=3\\ \Rightarrow\left\{{}\begin{matrix}b+d+c=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=4a\\a+b+c+d=4b\\a+b+c+d=4c\\a+b+c+d=4d\end{matrix}\right.\\ \Rightarrow4a=4b=4c=4d\Rightarrow a=b=c=d\\ \Rightarrow P=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
14 tháng 11 2021
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{a+b+d}=\dfrac{d}{a+b+c}=\dfrac{a+b+c+d}{3\left(a+b+c+d\right)}=\dfrac{1}{3}\\ \Rightarrow\left\{{}\begin{matrix}b+c+d=3a\\a+c+d=3b\\a+b+d=3c\\a+b+c=3d\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b+c+d=2a\\a+b+c+d=2b\\a+b+c+d=2c\\a+b+c+d=2d\end{matrix}\right.\\ \Rightarrow2a=2b=2c=2d\\ \Rightarrow a=b=c=d\\ \Rightarrow A=\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}+\dfrac{a+a}{a+a}=1+1+1+1=4\)
