Cho \(\frac{a}{c}=\frac{a-b}{b-c}\) (a;c không thể bằng 0,a không thể bằng b, b không thể bằng c) Chứng minh rằng \(\frac{1}{a}+\frac{1}{a-b}=\frac{1}{b-c}-\frac{1}{c}\)
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29 tháng 7 2020
Ta có: a + b + c = 0 => a + b = -c; b + c = -a; a + c = -b
a + b + c = 0 <=> a + b = -c
<=> (a + b)3 = (-c)3
<=> a3 + 3a2b + 3ab2 + b3 = -c3
<=> a3 + b3 + c3 = -3ab(a + b)
<=> a3 + b3 + c3 = 3abc (vì a + b = -c)
Khi đó: Q = \(\left(\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\right)\left(\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\right)\)
Q = \(1+\frac{a\left(a-b\right)}{c\left(b-c\right)}+\frac{b\left(a-b\right)}{c\left(c-a\right)}+\frac{c\left(b-c\right)}{a\left(a-b\right)}+1+\frac{b\left(b-c\right)}{a\left(c-a\right)}+\frac{c\left(c-a\right)}{b\left(a-b\right)}+\frac{a\left(c-a\right)}{b\left(b-c\right)}+1\)
Q = \(3+\left(\frac{a\left(a-b\right)}{c\left(b-c\right)}+\frac{a\left(c-a\right)}{b\left(b-c\right)}\right)+\left(\frac{b\left(a-b\right)}{c\left(c-a\right)}+\frac{b\left(b-c\right)}{a\left(c-a\right)}\right)+\left(\frac{c\left(b-c\right)}{a\left(a-b\right)}+\frac{c\left(c-a\right)}{b\left(a-b\right)}\right)\)
Q = \(3+\frac{ab\left(a-b\right)+ac\left(c-a\right)}{bc\left(b-c\right)}+\frac{ab\left(a-b\right)+bc\left(b-c\right)}{ac\left(c-a\right)}+\frac{bc\left(b-c\right)+ca\left(c-a\right)}{ab\left(a-b\right)}\)
Q = \(3+\frac{a\left(ab-b^2+c^2-ac\right)}{bc\left(b-c\right)}+\frac{b\left(a^2-ab+bc-c^2\right)}{ac\left(c-a\right)}+\frac{c\left(b^2-bc+ac-a^2\right)}{ab\left(a-b\right)}\)
Q = \(3+\frac{a\left[a\left(b-c\right)-\left(b-c\right)\left(b+c\right)\right]}{bc\left(b-c\right)}+\frac{b\left[b\left(c-a\right)-\left(c-a\right)\left(c+a\right)\right]}{ac\left(c-a\right)}+\frac{c\left[c\left(a-b\right)-\left(a-b\right)\left(a+b\right)\right]}{ab\left(a-b\right)}\)
Q = \(3+\frac{a\left[a-\left(b+c\right)\right]}{bc}+\frac{b\left(b-\left(c+a\right)\right)}{ac}+\frac{c\left[c-\left(a+b\right)\right]}{ab}\)
Q = \(3+\frac{a\left(a+a\right)}{bc}+\frac{b\left(b+b\right)}{ac}+\frac{c\left(c+c\right)}{ab}\)
Q = \(3+\frac{2a^2}{bc}+\frac{2b^2}{ac}+\frac{2c^2}{ab}\)
Q = \(3+\frac{2a^3+2b^3+2c^3}{abc}\)
Q = \(3+\frac{2\left(a^3+b^3+c^3\right)}{abc}\)
Q = \(3+\frac{2.3abc}{abc}=3+6=9\)
29 tháng 7 2020
Bài làm:
Đặt \(M=\frac{a-b}{c}+\frac{b-c}{a}+\frac{c-a}{b}\)
\(\Leftrightarrow abc.M=ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)\)
\(\Leftrightarrow abc.M=ab\left(a-b\right)+b^2c-bc^2+c^2a-ca^2\)
\(\Leftrightarrow abc.M=ab\left(a-b\right)+c^2\left(a-b\right)-c\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow abc.M=\left(a-b\right)\left(ab+c^2-ac-bc\right)\)
\(\Leftrightarrow abc.M=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)
\(\Rightarrow M=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}\)
Đặt \(N=\frac{c}{a-b}+\frac{a}{b-c}+\frac{b}{c-a}\)
\(\Rightarrow\left(a-b\right)\left(b-c\right)\left(c-a\right).N=c\left(b-c\right)\left(c-a\right)+a\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)\)
Mà \(a+b+c=0\Rightarrow\hept{\begin{cases}a=-b-c\\b=-c-a\\c=-a-b\end{cases}}\)
Thay vào ta được:
\(N=\frac{c\left(b-c\right)\left(c-a\right)-\left(b+c\right)\left(a-b\right)\left(c-a\right)+b\left(a-b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{c\left(c-a\right)\left(b-c-a+b\right)+b\left(a-b\right)\left(b-c-c+a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{c\left(c-a\right)\left(2b-c-a\right)+b\left(a-b\right)\left(a+b-2c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{c\left(c-a\right)\left(2b+b\right)+b\left(a-b\right)\left(-c-2c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{3bc\left(c-a\right)-3bc\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{3bc\left(b+c-2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(N=\frac{-9abc}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
Mà \(Q=M.N=\frac{\left(a-b\right)\left(a-c\right)\left(b-c\right)}{abc}.\frac{9abc}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=9\)
Vậy Q = 9
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 \(a=b=c\)
3 tháng 9 2020
Đặt \(\left(\frac{a-b}{c};\frac{b-c}{a};\frac{c-a}{b}\right)\rightarrow\left(x;y;z\right)\)
Khi đó:
\(S=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{x+z}{y}+\frac{y+z}{x}+\frac{x+y}{z}\)
Ta có:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-cb+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{\left(b-a\right)\left(b+a\right)-c\left(a-b\right)}{ab}\cdot\frac{c}{a-b}=\frac{\left(b-a\right)\left(b+a-c\right)}{ab}\cdot\frac{c}{a-b}=\frac{c\left(b+a-c\right)}{ab}\)
\(=\frac{2c^2}{ab}=\frac{2c^3}{abc}\)
Một cách tương tự khi đó:\(\frac{x+y}{z}+\frac{y+z}{x}+\frac{z+x}{y}=\frac{2\left(a^3+b^3+c^3\right)}{abc}=\frac{2\cdot3abc}{abc}=6\)
Khi đó:\(S=3+6=9\) Bạn để ý rằng \(a+b+c=0\) thì \(a^3+b^3+c^3=3abc\)
6 tháng 9 2020
sao \(\frac{c\left(b+a-c\right)}{ab}\) lại bằng \(\frac{2c^2}{ab}\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 12 2020
\(VT=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}\)
\(VT< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=2\)
\(VP=\dfrac{a}{\sqrt{a\left(b+c\right)}}+\dfrac{b}{\sqrt{b\left(c+a\right)}}+\dfrac{c}{\sqrt{c\left(a+b\right)}}\)
\(VP\ge\dfrac{2a}{a+b+c}+\dfrac{2b}{a+b+c}+\dfrac{2c}{a+b+c}=2\)
\(\Rightarrow VP>VT\) (đpcm)