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5 tháng 11 2017

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{-1}{c}\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=\dfrac{-1}{c^3}\) hay \(\dfrac{1}{a^3}+\dfrac{1}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\dfrac{1}{b^3}=\dfrac{-1}{c^3}\)

\(\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=\dfrac{3}{abc}\)

\(a^2b^2c^2.\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{3}{abc}.a^2b^2c^2\)

\(\Leftrightarrow\dfrac{b^2c^2}{a}+\dfrac{c^2a^2}{b}+\dfrac{a^2b^2}{c}=3abc\) hay\(M=3abc\left(đpcm\right)\)

30 tháng 10 2018

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=-\dfrac{1}{c}\)

\(\Rightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3=-\dfrac{1}{c^3}\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}\right)^3+\dfrac{1}{c^3}=0\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{3}{ab}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)=0\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}+\dfrac{3}{ab}.\left(-\dfrac{1}{c}\right)=0\)

\(\Rightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}-\dfrac{3}{abc}=0\Leftrightarrow\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}=\dfrac{3}{abc}\)

Ta có: Điều cần chứng minh là \(A=3abc\) hay \(\dfrac{A}{3abc}=1\)

Thật vậy:

\(\dfrac{A}{3abc}=\left(\dfrac{b^2c^2}{a}+\dfrac{c^2a^2}{b}+\dfrac{a^2b^2}{c}\right).\dfrac{1}{3abc}\)

\(\dfrac{A}{3abc}=\dfrac{b^2c^2}{3a^2bc}+\dfrac{c^2a^2}{3ab^2c}+\dfrac{a^2b^2}{3abc^2}\)

\(\dfrac{A}{3abc}=\dfrac{bc}{3a^2}+\dfrac{ac}{3b^2}+\dfrac{ab}{3c^2}\)

\(\dfrac{A}{3abc}=\dfrac{abc}{3a^3}+\dfrac{abc}{3b^3}+\dfrac{abc}{3c^3}\)

\(\dfrac{A}{3abc}=\dfrac{abc}{3}\left(\dfrac{1}{a^3}+\dfrac{1}{b^3}+\dfrac{1}{c^3}\right)=\dfrac{abc}{3}.\dfrac{3}{abc}=1\)

\(\dfrac{A}{3abc}=1\Leftrightarrow A=3abc\left(đpcm\right)\)

12 tháng 5 2017

a) Áp dụng bất đẳng thức Schur với \(r=1\)

\(\Rightarrow a^3+b^3+c^3+3abc\ge a^2b+ab^2+b^2c+bc^2+c^2a+ca^2\)

\(\Rightarrow3abc\ge a^2b+ca^2-a^3+ab^2+b^2c-b^3+c^2a+bc^2-c^3\)

\(\Rightarrow3abc\ge a^2\left(b+c-a\right)+b^2\left(a+c-b\right)+c^2\left(a+b-c\right)\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c\)

b) Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\dfrac{a^3}{b^2}+b+b\ge3\sqrt[3]{\dfrac{a^3}{b^2}.b^2}=3a\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{b^3}{c^2}+c+c\ge3b\\\dfrac{c^3}{a^2}+a+a\ge3c\end{matrix}\right.\)

\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}+2\left(a+b+c\right)\ge3\left(a+b+c\right)\)

\(\Rightarrow\dfrac{a^3}{b^2}+\dfrac{b^3}{c^2}+\dfrac{c^3}{a^2}\ge a+b+c\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c\)

c) Ta có \(abc=ab+bc+ca\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0

\(\Rightarrow\dfrac{1}{a+2b+3c}=\dfrac{1}{a+c+2\left(b+c\right)}\le\dfrac{1}{4}\left[\dfrac{1}{a+c}+\dfrac{1}{2\left(b+c\right)}\right]\)

Tương tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+2c+3a}\le\dfrac{1}{4}\left[\dfrac{1}{a+b}+\dfrac{1}{2\left(a+c\right)}\right]\\\dfrac{1}{c+2a+3b}\le\dfrac{1}{4}\left[\dfrac{1}{b+c}+\dfrac{1}{2\left(a+b\right)}\right]\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{1}{4}\left[\dfrac{3}{2}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\right]\)

\(\Rightarrow VT\le\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\) ( 1 )

Áp dụng bất đẳng thức \(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\) với a , b > 0

\(\Rightarrow\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{b+c}\le\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)\\\dfrac{1}{c+a}\le\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\end{matrix}\right.\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{4}\left(\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}\right)\right]\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{8}\left[\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\right]\)

\(\Rightarrow\dfrac{3}{8}\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\le\dfrac{3}{16}\) ( 2 )

Từ ( 1 ) và ( 2 )

\(\Rightarrow VT\le\dfrac{3}{16}\)

\(\Rightarrow\dfrac{1}{a+2b+3c}+\dfrac{1}{b+2c+3a}+\dfrac{1}{c+2a+3b}\le\dfrac{3}{16}\) ( đpcm )

12 tháng 5 2017

mk hỏi lâu rồi bây giờ bạn mới trả lời thì có đc GP k nhỉ

29 tháng 10 2018

\(A=\frac{b^2c^2}{a}+\frac{c^2a^2}{b}+\frac{a^2b^2}{c}=\frac{a^3b^3+b^3c^3+c^3a^3}{abc}=\frac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{abc}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{ab+bc+ca}{abc}=0\Rightarrow ab+bc+ca=0\)

\(\Rightarrow\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3=3.ab.bc.ca=3a^2b^2c^2\)

Vậy \(A=\frac{3a^2b^2c^2}{abc}=3abc\left(a,b,c\ne0\right)\)

25 tháng 5 2018

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

=> bc+ac+ab=0

ta có

\(bc+ac=-ab\)

<=> \(\left(bc+ac\right)^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)

tương tự

\(a^2b^2+b^2c^2-c^2a^2=-2ab^2c\)

\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)

thay vào E ta đc

\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)

=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)

14 tháng 1 2022
Cho sao nha nhưng tui ko bít làm
8 tháng 12 2018

Câu hỏi t/tự

17 tháng 9 2018

Hình như sai đề :

Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\)

\(\Leftrightarrow\dfrac{ab+ac+bc}{abc}=0\)

\(\Leftrightarrow ab+ac+bc=0\) ( do \(a;b;c\ne0\) ) ( 1 )

Từ ( 1 ) \(\Rightarrow ab+bc=-ac\)

\(\Rightarrow\left(ab+bc\right)^2=\left[-\left(ac\right)\right]^2\)

\(\Rightarrow a^2b^2+b^2c^2+2ab^2c=a^2c^2\) ( * )

CMTT , ta được : \(\left\{{}\begin{matrix}b^2c^2+c^2a^2+2bc^2a=a^2b^2\\c^2a^2+a^2b^2+2a^2cb=b^2c^2\end{matrix}\right.\) ( *' )

Thay ( * ) và ( * ') vào E , ta được :

\(E=\dfrac{a^2b^2c^2}{a^2b^2+b^2c^2-\left(a^2b^2+b^2c^2+2b^2ac\right)}+\dfrac{a^2b^2c^2}{b^2c^2+c^2a^2-\left(b^2c^2+c^2a^2+2bc^2a\right)}\)

\(+\dfrac{a^2b^2c^2}{c^2a^2+a^2b^2-\left(c^2a^2+a^2b^2+2a^2cb\right)}\)

\(=\dfrac{a^2b^2c^2}{-2b^2ac}+\dfrac{a^2b^2c^2}{-2c^2ab}+\dfrac{a^2b^2c^2}{-2a^2cb}\)

\(=\dfrac{-ac}{2}+\dfrac{-ab}{2}+\dfrac{-bc}{2}\)

\(=\dfrac{-\left(ac+ab+bc\right)}{2}\)

\(=\dfrac{0}{2}=0\)

Vậy \(E=0\)

NV
7 tháng 2 2022

\(VT=\dfrac{a^2}{b+ab^2c}+\dfrac{b^2}{b+abc^2}+\dfrac{c^2}{c+a^2bc}\ge\dfrac{\left(a+b+c\right)^2}{a+b+c+abc\left(a+b+c\right)}=\dfrac{9}{3+3abc}\)

\(VT\ge\dfrac{9}{3+\dfrac{\left(a+b+c\right)^3}{9}}=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

8 tháng 2 2022

cảm ơn thầy ạ

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: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)

\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)

\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)

\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)

\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)

\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)

 

 

23 tháng 11 2021

Cho a,b,c >0, chứng minh rằng :\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{... - Hoc24