Cho 3 số a,b,c khác 0 thỏa mãn a+b+c=0.
Tính \(P=\frac{a}{c}\cdot\frac{a^2-b^2-c^2}{b^2-c^2-a^2}\cdot\frac{c^2+a^2-b^2}{c^2-a^2-b^2}\)
HELP Tớ với ạ.Thanks trước:v
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10 tháng 8 2016
Bài 1 :
a) Ta có : \(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng bđt Cauchy : \(a+b\ge2\sqrt{ab}\) , \(b+c\ge2\sqrt{bc}\) , \(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) hay \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge8abc\)
30 tháng 8 2019
Đặt \(\left(\frac{a-b}{c},\frac{b-c}{a},\frac{c-a}{b}\right)\rightarrow\left(x,y,z\right)\)
Khi đó:\(\left(\frac{c}{a-b},\frac{a}{b-c},\frac{b}{c-a}\right)\rightarrow\left(\frac{1}{x},\frac{1}{y},\frac{1}{z}\right)\)
Ta có:
\(P\cdot Q=\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3+\frac{y+z}{x}+\frac{z+x}{y}+\frac{x+y}{z}\)
Mặt khác:\(\frac{y+z}{x}=\left(\frac{b-c}{a}+\frac{c-a}{b}\right)\cdot\frac{c}{a-b}=\frac{b^2-bc+ac-a^2}{ab}\cdot\frac{c}{a-b}\)
\(=\frac{c\left(a-b\right)\left(c-a-b\right)}{ab\left(a-b\right)}=\frac{c\left(c-a-b\right)}{ab}=\frac{2c^2}{ab}\left(1\right)\)
Tương tự:\(\frac{x+z}{y}=\frac{2a^2}{bc}\left(2\right)\)
\(=\frac{x+y}{z}=\frac{2b^2}{ac}\left(3\right)\)
Từ ( 1 );( 2 );( 3 ) ta có:
\(P\cdot Q=3+\frac{2c^2}{ab}+\frac{2a^2}{bc}+\frac{2b^2}{ac}=3+\frac{2}{abc}\left(a^3+b^3+c^3\right)\)
Ta có:\(a+b+c=0\)
\(\Rightarrow\left(a+b\right)^3=-c^3\)
\(\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)
\(\Rightarrow a^3+b^3+c^3=3abc\)
Khi đó:\(P\cdot Q=3+\frac{2}{abc}\cdot3abc=9\)
NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 10 2020
\(ab+bc+ca=3abc\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\)
\(Q=\frac{a^2+c^2-c^2}{a\left(c^2+a^2\right)}+\frac{b^2+a^2-a^2}{a\left(a^2+b^2\right)}+\frac{c^2+b^2-b^2}{b\left(b^2+c^2\right)}\)
\(Q=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\left(\frac{a}{a^2+b^2}+\frac{b}{b^2+c^2}+\frac{c}{c^2+a^2}\right)\)
\(Q\ge3-\left(\frac{a}{2ab}+\frac{b}{2bc}+\frac{c}{2ca}\right)=3-\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{3}{2}\)
\(Q_{min}=\frac{3}{2}\) khi \(a=b=c=1\)
MP
4 tháng 7 2020
Áp dụng BĐT Cauchy, ta có :
\(a^2+b^2\ge2ab\)
\(b^2+1\ge2b\)
\(\Rightarrow\) \(a^2+2b^2+3\ge2\left(ab+b+1\right)\)
\(\Rightarrow\) \(\frac{1}{a^2+2b^2+3}\le\frac{1}{2\left(ab+b+1\right)}\) ( 1 )
Tương tự : \(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\) ( 2 )
\(\frac{1}{c^2+2a^2+3}\le\frac{1}{2\left(ac+a+1\right)}\) ( 3 )
Từ ( 1 ), ( 2 ) và ( 3 ) cộng vế theo vế, ta có :
\(VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\right)\)
Đặt \(A=\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}=\frac{ac}{ab.ac+abc+ac}+\frac{a}{abc+ac+a}+\frac{1}{ac+a+1}\)
\(=\frac{ac+a+1}{ac+a+1}=1\)
\(\Rightarrow\) \(VT\le\frac{1}{2}.1=\frac{1}{2}\)
\(\Rightarrow\) đpcm
17 tháng 8 2016
Đề đúng : Cho a,b,c > 0 và \(a+b+c\le1\)
CMR : \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge9\)
Đặt \(x=a^2+2bc,y=b^2+2ac,z=c^2+2ab\)
Áp dụng bđt Bunhiacopxki , ta có: \(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\left(x+y+z\right)\ge\left(\sqrt{\frac{1}{x}.x}+\sqrt{\frac{1}{y}.y}+\sqrt{\frac{1}{z}.z}\right)^2=9\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\) hay \(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{9}{\left(a+b+c\right)^2}\ge9\)
17 tháng 8 2016
Ta thấy: \(\left(a^2+2bc\right)+\left(b^2+2ac\right)+\left(c^2+2ab\right)=\left(a+b+c\right)^2\le1\)
Sử dụng Cosi 3 số ta suy ra
\(VT\ge\left[\left(a^2+2bc\right)+\left(b^2+2ac\right)+\left(c^2+2ab\right)\right]\left(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\right)\)
\(\ge3\sqrt[3]{\left(a^2+2bc\right)\left(b^2+2ac\right)\left(c^2+2ab\right)}\cdot3\sqrt[3]{\frac{1}{a^2+2bc}\cdot\frac{1}{b^2+2ac}\cdot\frac{1}{c^2+2ab}}=9\) (Đpcm)
Đẳng thức xảy ra khi\(\begin{cases}a+b+c=1\\a^2+2bc=b^2+2ac=c^2+2ab\end{cases}\)\(\Leftrightarrow a=b=c=\frac{1}{3}\)
4 tháng 11 2018
\(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+2abc=0\)
\(\Rightarrow ab^2+ac^2+bc^2+ba^2+c\left(a+b\right)^2=0\)
\(\Rightarrow ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2=0\)
\(\Rightarrow\left(a+b\right)\left(ab+c^2+ca+cb\right)=0\)
\(\Rightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
Từ đó a = -b hoặc b = -c hoặc c = -a
Nếu a = -b mà \(a^3+b^3+c^3=1\Rightarrow\left(-b\right)^3+b^3+c^3=1\Rightarrow c^3=1\Rightarrow c=1\)
Khi đó: \(A=\frac{1}{\left(-b\right)^{2017}}+\frac{1}{b^{2017}}+\frac{1}{1^{2017}}=0+1=1\)
Tương tự với các trường hợp b = -c và a = -c, ta tính được A = 1
a+b+c=0=> a2-b2-c2=2bc,b2-c2-a2=2ac,c2+a2-b2=-2ac,c2-a2-b2=2ab
=>\(P=\frac{a}{c}.\frac{2bc}{2ac}.\frac{-2ac}{2ab}=-1\)
a+b+c=0 <=> a+b=-c; b+c=-a;c+a=-b
\(\frac{a^2-b^2-c^2}{b^2-c^2-a^2}=\frac{\left(a-c\right)\left(a+c\right)-b^2}{\left(b-a\right)\left(b+a\right)-c^2}=\frac{\left(a-c\right)\left(-b\right)-b^2}{\left(b-a\right)\left(-c\right)-c^2}=\frac{b\left(c-a-b\right)}{c\left(a-b-c\right)}\)
\(=\frac{b\left[c-\left(a+b\right)\right]}{c\left[a-\left(b+c\right)\right]}=\frac{b\left[c-\left(-c\right)\right]}{c\left[a-\left(-a\right)\right]}=\frac{b.2c}{c.2a}=\frac{b}{a}\)
***
\(\frac{c^2+a^2-b^2}{c^2-a^2-b^2}=\frac{\left(c-b\right)\left(c+b\right)+a^2}{\left(c-b\right)\left(c+b\right)-a^2}=\frac{\left(c-b\right)\left(-a\right)+a^2}{\left(c-b\right)\left(-a\right)-a^2}=\frac{a\left(a+b-c\right)}{a\left(b-c-a\right)}\)
\(=\frac{a+b-c}{b-\left(c+a\right)}=\frac{-c-c}{b-\left(-b\right)}=\frac{-2c}{2b}=\frac{-c}{b}\)
\(P=\frac{a}{c}.\frac{a^2-b^2-c^2}{b^2-c^2-a^2}.\frac{c^2+a^2-b^2}{c^2-a^2-b^2}=\frac{a}{c}.\frac{b}{a}.\frac{-c}{b}=-1\)