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Lời giải:
\(F=xyz-(x+y+z)=\frac{b+c}{a}.\frac{c+a}{b}.\frac{a+b}{c}-\left(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}\right)\)
\(=\frac{(b+c)(c+a)(a+b)}{abc}-\frac{bc(b+c)+ca(c+a)+ab(a+b)}{abc}\)
\(=\frac{ab(a+b)+bc(b+c)+ca(c+a)+2abc}{abc}-\frac{bc(b+c)+ca(c+a)+ab(a+b)}{abc}\)
\(=\frac{2abc}{abc}=2\)
vì có 1 chút nhầm lẫn nên giờ mk mới ra mong bạn thứ lỗi
bài 1
\(\Leftrightarrow\frac{4a^4}{2a^3+2a^2b^2}+\frac{4b^4}{2b^3+2c^2b^2}+\frac{4c^4}{2c^3+2a^2c^2}\)
\(\ge\frac{\left(2a^2+2b^2+2c^2\right)^2}{2a^3+2b^3+2c^3+2a^2b^2+2c^2b^2+2a^2c^2}\)
\(\ge\frac{36}{a^4+a^2+b^4+b^2+c^4+c^2+2a^2b^2+2c^2b^2+2a^2c^2}\)
\(=\frac{36}{\left(a^2+b^2+c^2\right)^2+a^2+b^2+c^2}=3\ge a+b+c\)
Dấu bằng xảy ra khi \(a=b=c=1\)
Bài 2 là chuyên Bình Thuận, 2016-2017
Áp dụng bất đẳng thức Cauchy – Schwarz, ta có:
\(\frac{xy}{x^2+yz+zx}\le\frac{xy\left(y^2+yz+zx\right)}{\left(x^2+yz+zx\right)\left(y^2+yz+zx\right)}\le\frac{xy\left(y^2+yz+zx\right)}{\left(xy+yz+zx\right)^2}\)
Tương tự: \(\frac{yz}{y^2+zx+xy}\le\frac{xy\left(z^2+zx+xy\right)}{\left(xy+yz+zx\right)^2}\);\(\frac{zx}{z^2+xy+yz}\le\frac{zx\left(x^2+xy+yz\right)}{\left(xy+yz+zx\right)^2}\)
Cộng từng vế của 3 BĐT trên. ta được:
\(VT\le\frac{\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)}{\left(xy+yz+zx\right)^2}=\frac{x^2+y^2+z^2}{xy+yz+zx}\)
Đẳng thức xảy ra khi x = y = z
Xét: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}=\frac{bc\left(b-c\right)+ca\left(c-a\right)+ab\left(a-b\right)}{abc}\)
\(=\frac{b^2c-bc^2+ca\left(c-a\right)+a^2b-ab^2}{abc}=\frac{b^2\left(c-a\right)+ca\left(c-a\right)-b\left(c^2-a^2\right)}{abc}\)
\(=\frac{\left(c-a\right)\left(b^2+ca\right)-b\left(c-a\right)\left(c+a\right)}{abc}=\frac{\left(c-a\right)\left(b^2+ca-bc-ba\right)}{abc}\)
\(=\frac{\left(c-a\right)\left(b-a\right)\left(b-c\right)}{abc}=-\frac{\left(b-c\right)\left(c-a\right)\left(a-b\right)}{abc}=-\frac{1}{xyz}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{-1}{xyz}\Leftrightarrow xy+yz+zx=-1\)
\(xy+yz+zx=\frac{a}{b-c}.\frac{b}{c-a}+\frac{b}{c-a}.\frac{c}{a-b}+\frac{c}{a-b}.\frac{a}{b-c}\)\(=\frac{ab\left(a-b\right)+bc\left(b-c\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{a^2b-ab^2+b^2c-bc^2+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{b\left(a^2-c^2\right)+b^2\left(c-a\right)+ca\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)
\(=\frac{\left(c-a\right)\left(b^2+ca-ab-bc\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(c-a\right)\left(b\left(b-a\right)+c\left(a-b\right)\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(a-b\right)\left(c-b\right)\left(c-a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)
Giải
P = \(\frac{x}{xy+3z}+\frac{y}{yz+3z}+\frac{z}{zx+3x}\)\(=\frac{x}{xy+\left(x+y+z\right)z}+\frac{y}{yz+\left(x+y+z\right)x}+\frac{z}{zx+\left(x+y+z\right)y}\)
\(=\frac{x}{\left(x+z\right)\left(y+z\right)}+\frac{y}{\left(x+y\right)\left(x+z\right)}+\frac{z}{\left(x+y\right)\left(y+z\right)}\)\(=\frac{x\left(x+y\right)+y\left(y+z\right)+z\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)=\(=\frac{x^2+y^2+z^2+xy+yz+zx}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\frac{\left(x+y+z\right)^2-\left(xy+yz+xz\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
Theo BĐT CÔSI : \(\left(x+y\right)\left(y+z\right)\left(z+x\right)\le\frac{\left(2x+2y+2z\right)^3}{27}=8\)
\(xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=3\)
Do Đó : \(P\ge\frac{3^2-3}{8}=\frac{2}{3}\)
Vậy Min P= 2/3 dấu = <=> x=y=z=1
tik cho mik nha !!!!
Hạ sách : Nhân hết ra :)))
Ta có :
\(A=\left(x+\frac{1}{x}\right)^2+\left(y+\frac{1}{y}\right)^2+\left(xy+\frac{1}{xy}\right)^2-\left(x+\frac{1}{x}\right)\left(y+\frac{1}{y}\right)\left(xy+\frac{1}{xy}\right)\)
\(=x^2+\frac{1}{x^2}+2+y^2+\frac{1}{y^2}+2+x^2y^2+\frac{1}{x^2y^2}+2-\left(xy+\frac{x}{y}+\frac{y}{x}+\frac{1}{xy}\right)\left(xy+\frac{1}{xy}\right)\)
\(=x^2+y^2+\frac{1}{x^2y^2}+x^2y^2+\frac{1}{x^2}+\frac{1}{y^2}+6-\left(x^2y^2+1+x^2+\frac{1}{y^2}+y^2+\frac{1}{x^2}+1+\frac{1}{x^2y^2}\right)\)
\(=6-1-1\)
\(=4\)
Lời giải:
Với điều kiện đã cho ta có:
\(F=xy+yz+xz+2xyz=\frac{ab}{(b+c)(c+a)}+\frac{bc}{(c+a)(a+b)}+\frac{ac}{(b+c)(a+b)}+\frac{2abc}{(b+c)(c+a)(a+b)}\)
\(=\frac{ab(a+b)+bc(b+c)+ac(a+c)+2abc}{(a+b)(b+c)(c+a)}\)
\(=\frac{ab(a+b+c)+bc(b+c+a)+ac(a+c)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+b+c)(ab+bc)+ac(a+c)}{(a+b)(b+c)(c+a)}=\frac{(a+c)(ba+b^2+bc+ac)}{(a+b)(b+c)(c+a)}\)
\(=\frac{(a+c)[b(a+b)+c(b+a)]}{(a+b)(b+c)(c+a)}=\frac{(a+c)(b+c)(a+b)}{(a+b)(b+c)(c+a)}=1\)