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\(\frac{1}{a}+\frac{1}{b}-\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow abc\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-2abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
CHÚC BẠN HỌC TỐT
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\Leftrightarrow\frac{bc+ac-ab}{abc}=0\)
Vì \(a,b,c\ne0\Rightarrow a.b.c\ne0\)
\(\Rightarrow bc+ac-ab=0\)
\(\Rightarrow\hept{\begin{cases}\left(bc+ac\right)^2=\left(ab\right)^2\\\left(bc-ab\right)^2=\left(-ac\right)^2\\\left(ac-ab\right)^2=\left(-bc\right)^2\end{cases}\Rightarrow}\hept{\begin{cases}b^2c^2+c^2a^2-a^2b^2=-abc^2\\b^2c^2+a^2b^2-a^2c^2=2ab^2c\\a^2c^2+a^2b^2-b^2c^2=2a^2bc\end{cases}}\)
\(\Rightarrow E=\frac{a^2b^2c^2}{2ab^2c}+\frac{a^2b^2c^2}{-2abc^2}+\frac{a^2b^2c^2}{2a^2bc}\)
\(\Rightarrow E=\frac{ac}{2}-\frac{ab}{2}+\frac{bc}{2}=\frac{ac-ab+bc}{2}=\frac{0}{2}=0\)
Vậy \(E=0\)
Anh/chị tham khảo ở đây nhé :
đặt x=a^2 + 2bc, y=b^2 + 2ac, z=c^2 + 2ab
=> x + y + z = (a + b + c)^2 <(=) 1
VT bpt : 1/x + 1/y + 1/z >(=) 3.căn3(1/xyz)...dùng cô-si cho 3 số nhé
mà x + y + z >(=) 3.căn3(xyz) <(=) 1
<=> 1/( 3.căn3 (xyz) >(=) 1 (ở đây là đổi chiều bđt)
<=> 1/ căn3 (xyz) >(=) 3
=> VT: 1/x + 1/y + 1/z >(=) 3.3 = 9
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\) <=> \(\frac{1}{a}+\frac{1}{b}=-\frac{1}{c}\)
<=> \(\left(\frac{1}{a}+\frac{1}{b}\right)^3=\left(-\frac{1}{c}\right)^3\)
<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{3}{a^2b}+\frac{3}{ab^2}=-\frac{1}{c^3}\)
<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=-\frac{3}{ab}\left(\frac{1}{a}+\frac{1}{b}\right)\)
<=> \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
Khi đó, A = \(\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)=abc\cdot\frac{3}{abc}=3\)
Xét: \(A=\frac{bc}{a^2}+\frac{ac}{b^2}+\frac{ab}{c^2}=abc\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
Ta có đẳng thức sau: \(x^3+y^3+z^3=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)+3xyz\)
(Đẳng thức này chứng minh rất dễ nha, chỉ cần bung hết ra là được)
Vậy ta thế \(x=\frac{1}{a},y=\frac{1}{b},z=\frac{1}{c}\)vào đẳng thức:
\(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}-\frac{1}{ab}-\frac{1}{bc}-\frac{1}{ca}\right)+\frac{3}{abc}\)
\(=\frac{3}{abc}\)Vì \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)---> Thế cái này vào A:
\(\Rightarrow A=abc.\frac{3}{abc}=3\)
Xoooooooong !!!!! :)))
Nhân cả 2 vế với a+b+c
Chứng minh \(\frac{a}{b}+\frac{b}{a}\ge2\) tương tự với \(\frac{b}{c}+\frac{c}{b};\frac{c}{a}+\frac{a}{c}\)
\(\Leftrightarrow\frac{a}{b}+\frac{b}{a}-2\ge0\Leftrightarrow\frac{a^2-2ab+b^2}{ab}\ge0\Leftrightarrow\frac{\left(a-b\right)^2}{ab}\ge0\)luôn đúng do a;b>0
dễ rồi nhé
b) \(P=\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\)
\(P=\left(\frac{x+1}{x+1}+\frac{y+1}{y+1}+\frac{z+1}{z+1}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=\left(1+1+1\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng bđt Cauchy Schwarz dạng Engel (mình nói bđt như vậy,chỗ này bạn cứ nói theo cái bđt đề bài cho đi) ta được:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\ge\frac{\left(1+1+1\right)^2}{x+1+y+1+z+1}=\frac{9}{4}\)
=>\(P=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\le3-\frac{9}{4}=\frac{3}{4}\)
=>Pmax=3/4 <=> x=y=z=1/3
1. Ta có : x + y + z = 0 \(\Rightarrow\)( x + y + z )2 = 0 \(\Rightarrow\)x2 + y2 + z2 = - 2 ( xy + yz + xz )\(S=\frac{x^2+y^2+z^2}{\left(y-z\right)^2+\left(z-x\right)^2+\left(x-y\right)^2}=\frac{-2\left(xy+yz+xz\right)}{2\left(x^2+y^2+z^2\right)-2\left(yz+xz+xy\right)}\)
\(S=\frac{-2\left(xy+yz+xz\right)}{-4\left(xy+yz+xz\right)-2\left(yz+xz+xy\right)}=\frac{-2\left(xy+yz+xz\right)}{-6\left(xy+yz+xz\right)}=\frac{1}{3}\)
Bài 1:Cách thông thường nhất là sos hoặc cauchy-Schwarz nhưng thôi ko làm:v Thử cách này cho nó mới dù rằng ko chắc
Giả sử \(a\ge b\ge c\Rightarrow c\le1\Rightarrow a+b=3-c\ge2\) và \(a\ge1\)
Ta có \(LHS=a^3.a+b^3.b+c^3.c\)
\(=\left(a^3-b^3\right)a+\left(b^3-c^3\right)\left(a+b\right)+c^3\left(a+b+c\right)\)
\(\ge\left(a^3-b^3\right).1+\left(b^3-c^3\right).2+3c^3\)
\(=a^3+b^3+c^3=RHS\)
Đẳng thức xảy ra khi a = b = c = 1
abc=a+b+c => 1 = 1/ab + 1/bc + 1/ac
2 = 1/a+1/b+1/c => 4 = 1/a^2 + 1/b^2 + 1/c^2 + 2/ab + 2/ac + 2/cb
=> 4 = 1/a^2 + 1/b^2 + 1/c^2 + 2(1/ab + 1/ac + 1/bc) = M + 2
=> M = 4 - 2 = 2
Mk làm bài đầu thôi,sáng nay mk làm cái tt cho
\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\)
\(\Leftrightarrow\)\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)
\(\Leftrightarrow\)\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{ab}+\frac{2}{bc}+\frac{2}{ca}=4\)
\(\Leftrightarrow\)\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{c}{abc}+\frac{a}{abc}+\frac{b}{abc}\right)=4\)
\(\Leftrightarrow\)\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\frac{a+b+c}{abc}=4\)
\(\Leftrightarrow\)\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\) (do a+b+c = abc)
\(\Leftrightarrow\)\(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2bc+2ac=0\)
\(\Leftrightarrow a^2+b^2-c^2=-2c^2-2bc-2ac-2ab\)
\(\Leftrightarrow a^2+b^2-c^2=-\left[2c.\left(c+b\right)+2a.\left(c+b\right)\right]\)
\(\Leftrightarrow a^2+b^2-c^2=-2.\left(a+c\right)\left(c+b\right)\)
Tương tự \(b^2+c^2-a^2=-2.\left(a+b\right)\left(a+c\right)\)
\(c^2+a^2-b^2=-2.\left(b+c\right)\left(b+a\right)\)
Đặt \(A=\frac{1}{a^2+b^2-c^2}+\frac{1}{b^2+c^2-a^2}+\frac{1}{c^2+a^2-b^2}\)
\(=-\frac{1}{2}.\left[\frac{1}{\left(b+c\right)\left(a+c\right)}+\frac{1}{\left(a+b\right)\left(a+c\right)}+\frac{1}{\left(b+c\right)\left(a+b\right)}\right]\)
\(=-\frac{1}{2}.\frac{a+b+b+c+a+c}{\left(b+c\right).\left(a+c\right)\left(a+b\right)}=-\frac{1}{2}.\frac{2.\left(a+b+c\right)}{\left(b+c\right).\left(a+c\right).\left(a+b\right)}=0\)