Cho a,b,c đôi một khác nhau cmr (a+b)^2÷(a-b)^2 + (b+c)^2÷(b-c)^2 + (c+a)^2÷(c-a)^2>=2
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20 tháng 12 2020
BĐT đã cho tương đương với:
\(\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)^2-2\left[\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}+\dfrac{ca}{\left(a-b\right)\left(b-c\right)}\right]\ge2\left(\cdot\right)\).
Mặt khác ta có: \(\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}+\dfrac{ca}{\left(a-b\right)\left(b-c\right)}=\dfrac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{-\left(a-b\right)\left(b-c\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\).
Do đó \(\left(\cdot\right)\Leftrightarrow\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)^2\ge0\) (luôn đúng).
BĐT đã cho dc c/m.
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 1 2024
Trước hết ta có:
\(\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{ac}{\left(b-c\right)\left(a-b\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{ab\left(a-b\right)+b^2c-a^2c+ac^2-bc^2}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{\left(a-b\right)\left(ab-ac-bc+c^2\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\dfrac{\left(a-b\right)\left[a\left(b-c\right)-c\left(b-c\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)
Do đó:
\(\left(\dfrac{a}{b-c}\right)^2+\left(\dfrac{b}{c-a}\right)^2+\left(\dfrac{c}{a-b}\right)^2-2+2\)
\(=\left(\dfrac{a}{b-c}\right)^2+\left(\dfrac{b}{c-a}\right)^2+\left(\dfrac{c}{a-b}\right)^2+2\left(\dfrac{ab}{\left(b-c\right)\left(c-a\right)}+\dfrac{ac}{\left(a-b\right)\left(b-c\right)}+\dfrac{bc}{\left(c-a\right)\left(a-b\right)}\right)+2\)
\(=\left(\dfrac{a}{b-c}+\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)^2+2\ge2\) (đpcm)
25 tháng 6 2021
Ta có: \(a^2-b=b^2-c\Leftrightarrow a^2-b^2=b-c\)
\(\Leftrightarrow\left(a-b\right)\left(a+b\right)=b-c\Rightarrow a+b=\frac{b-c}{a-b}\)
Tương tự CM được: \(b+c=\frac{c-a}{b-c}\) và \(c+a=\frac{a-b}{c-a}\)
Khi đó:
\(\left(a+b+1\right)\left(b+c+1\right)\left(c+a+1\right)\)
\(=\left(\frac{a-b}{c-a}+1\right)\left(\frac{c-a}{b-c}+1\right)\left(\frac{b-c}{a-b}+1\right)\)
\(=\frac{c-b}{c-a}\cdot\frac{b-a}{b-c}\cdot\frac{a-c}{a-b}=-1\)