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Câu 1:
- Chứng minh a3+b3+c3=3abc thì a+b+c=0
\(a^3+b^3+c^3=3abc\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b\right)^3-3a^2b-3ab^2+c^3-3abc=0\)
\(\Rightarrow\left[\left(a+b\right)^3+c^3\right]-3abc\left(a+b+c\right)=0\)
\(\Rightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Rightarrow0=0\) Đúng (Đpcm)
- Chứng minh a3+b3+c3=3abc thì a=b=c
Áp dụng Bđt Cô si 3 số ta có:
\(a^3+b^3+c^3\ge3\sqrt[3]{a^3b^3c^3}=3abc\)
Dấu = khi a=b=c (Đpcm)
Câu 2
Từ \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Rightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=3\cdot\frac{1}{abc}\)
Ta có:
\(\frac{ab}{c^2}+\frac{bc}{a^2}+\frac{ac}{b^2}=\frac{abc}{c^3}+\frac{abc}{a^3}+\frac{abc}{b^3}\)
\(=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc\cdot3\cdot\frac{1}{abc}=3\)
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 \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge8abc\)
1/a+1/b+1/c = 0 <=>ab+bc+ca/abc=0
=> ab+bc+ca = 0
Khi đó : (a+b+c)^2 = a^2+b^2+c^2+2.(ab+bc+ca) = a^2+b^2+c^2+2.0 = a^2+b^2+c^2
=> ĐPCM
k mk nha
\(a+b+c=0\Leftrightarrow\frac{a+b+c}{abc}=0\Leftrightarrow\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=0\)
\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)
Bài làm :
Ta có :
\(\left(a+b+c\right)^2=a^2+b^2+c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=a^2+b^2+c^2\)
\(\Leftrightarrow2ab+2bc+2ac=0\)
\(\Leftrightarrow2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow ab+bc+ac=0\)
\(\Leftrightarrow\frac{ab+bc+ac}{abc}=0\)
\(\Leftrightarrow\frac{ab}{abc}+\frac{bc}{abc}+\frac{ac}{abc}=0\)
\(\Leftrightarrow\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=0\)
\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\left(1\right)\)
\(\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)=-\frac{1}{c^3}\left(2\right)\)
Thay (1) vào (2) ; ta được :
\(\frac{1}{a^3}+\frac{1}{b^3}-\frac{3}{abc}=-\frac{1}{c^3}\)
\(\Leftrightarrow\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
=> Điều phải chứng minh
Ta có \(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc=a^2+b^2+c^2\)
\(\Leftrightarrow2ab+2ac+2bc=0\)
\(\Leftrightarrow2\left(ab+ac+bc\right)=0\)
\(\Leftrightarrow ab+ac+bc=0\)
Ta lại có giả sử
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(\Leftrightarrow\frac{a^3b^3+b^3c^3+c^3a^3}{a^3b^3c^3}=\frac{3}{abc}\)
\(\Leftrightarrow\frac{a^3b^3+b^3c^3+c^3a^3}{a^2b^2c^2}=3\)
\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3=3.a^2b^2c^2\)
\(\Leftrightarrow a^3b^3+b^3c^3+c^3a^3-3.a^2b^2c^2=0\)
\(\Leftrightarrow\left(ab+bc+ac\right)^3-3ca\left(ab+bc\right)\left(ab+bc+ac\right)-3ab^3c\left(-ac\right)-3a^2b^2c^2=0\)
\(\Leftrightarrow0+3a^2b^2c^2-3a^2b^2c^2+0=0\)
\(\Leftrightarrow0=0\left(lđ\right)\)
Vậy bất đẳng thức được chứng minh
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)
\(\frac{yz}{xyz}+\frac{xz}{xyz}+\frac{xy}{xyz}=0\)
\(\frac{yz+xz+xy}{xyz}=0\)
yz + xz + xy = 0
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2\times\left(xy+xz+yz\right)=x^2+y^2+z^2+2\times0=x^2+y^2+z^2\left(\text{đ}pcm\right)\)
a) Từ giả thiết suy ra: xy + yz + zx = 0
Do đó:
\(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)=x^2+y^2+z^2\)
b) Đặt \(\frac{1}{a-b}=x\); \(\frac{1}{b-c}=y\); \(\frac{1}{c-a}=z\)
Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=a-b+b-c+c-a=0\)
Theo câu a ta có: \(x^2+y^2+z^2=\left(x+y+z\right)^2\)
Suy ra điều phải chứng minh
1) Áp dụng bunhiacopxki ta được \(\sqrt{\left(2a^2+b^2\right)\left(2a^2+c^2\right)}\ge\sqrt{\left(2a^2+bc\right)^2}=2a^2+bc\), tương tự với các mẫu ta được vế trái \(\le\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}\le1< =>\)\(1-\frac{bc}{2a^2+bc}+1-\frac{ac}{2b^2+ac}+1-\frac{ab}{2c^2+ab}\le2< =>\)
\(\frac{bc}{2a^2+bc}+\frac{ac}{2b^2+ac}+\frac{ab}{2c^2+ab}\ge1\)<=> \(\frac{b^2c^2}{2a^2bc+b^2c^2}+\frac{a^2c^2}{2b^2ac+a^2c^2}+\frac{a^2b^2}{2c^2ab+a^2b^2}\ge1\) (1)
áp dụng (x2 +y2 +z2)(m2+n2+p2) \(\ge\left(xm+yn+zp\right)^2\)
(2a2bc +b2c2 + 2b2ac+a2c2 + 2c2ab+a2b2). VT\(\ge\left(bc+ca+ab\right)^2\) <=> (ab+bc+ca)2. VT \(\ge\left(ab+bc+ca\right)^2< =>VT\ge1\) ( vậy (1) đúng)
dấu '=' khi a=b=c
Ta co: a+b+c=0
=>a+b=-c
=>c2=a2+b2+2ab
=>a2+b2-c2=-2ab
=>\(\frac{1}{a^2+b^2-c^2}=\frac{1}{-2ab}\)
Tuong tu ....
=> \(\frac{1}{a^2+b^2+c^2}+...\)=\(\frac{1}{-2ab}+\frac{1}{-2bc}+\frac{1}{-2ac}\)
=\(\frac{a+b+c}{-2abc}\)
=0(ĐPCM)