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\(VT=\left(\frac{1}{x^3+y^3+xy\left(x+y\right)}+\frac{1}{2xy}\right)+\left(\frac{1}{4xy}+4xy\right)+\frac{5}{4xy}\)
\(\ge\frac{4}{x^3+y^3+xy\left(x+y\right)+2xy\left(x+y\right)}+2+\frac{5}{\left(x+y\right)^2}=11\)
Đẳng thức xảy ra khi \(x=y=\frac{1}{2}\)
Ta có:
\(\left(a+b\right)^2\ge4ab\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{1}{a+b}\) với a,b dương
Do x+y=1 nên ta có:
\(A=\frac{1}{x^3+xy+y^3}+\frac{4y^2x^2+2}{xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(4xy+\frac{1}{4xy}\right)+\frac{5}{4xy}\)
Ta có:
\(\frac{1}{x^2+y^2}+\frac{1}{2xy}\ge\frac{4}{\left(x+y\right)^2}=4\)
Ta sử dung bđt \(\frac{a}{b}+\frac{b}{a}\ge2\left(a,b>0\right)\)thì \(4xy+\frac{1}{4xy}=\frac{4xy}{1}+\frac{1}{4xy}\ge2\)
Mặt khác
\(1=\left(x+y\right)^2\ge4xy\Rightarrow xy\le\frac{1}{4}\Rightarrow\frac{5}{4xy}\ge5\)Nên ta suy ra:
\(A=\frac{1}{x^3+xy+y^3}+\frac{4y^2x^2+2}{xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\left(4xy+\frac{1}{4xy}\right)+\frac{5}{4xy}\ge4+2+5=11\)
Dấu "=" xảy ra khi và chỉ khi x=y=\(\frac{1}{2}\)
sửa đề CM biểu thức \(\le\frac{3}{16}\)
\(\frac{1}{2x+y+z}=\frac{1}{x+x+y+z}\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
bình phương hai vế:
\(\frac{1}{2x+y+z)^2}=\frac{1}{16^2}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
áp dụng bđt phụ: \(\left(x_1+x_2+x_3+x_4\right)^2\le4\left(x_1+x_2+x_3+x_4\right)\)
áp dụng cho cụm \(\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\)
=> \(\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\le4\left(\frac{2}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
=> \(\frac{1}{\left(2x+y+z\right)^2}\le\frac{1}{64}\left(\frac{2}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
áp dụng tương tự:
=> \(\frac{1}{\left(2y+x+z\right)}\le\frac{1}{64}\left(\frac{1}{x^2}+\frac{2}{y^2}+\frac{1}{z^2}\right)\)
\(\frac{1}{\left(2z+x+y\right)^2}\le\frac{1}{64}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{z^2}\right)\)
cộng cả ba biêu thức trên
=> \(VT\le\frac{1}{64}\left\lbrace\left(\frac{2}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+\left(\frac{1}{x^2}+\frac{2}{y^2}+\frac{1}{z^2}\right)+\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{z^2}\right)\right\rbrace\) \(VT\le\frac{1}{64}\left(\frac{4}{x^2}+\frac{4}{y^2}+\frac{4}{z^2}\right)\)
\(VT\le\frac{4}{64}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)=\frac{1}{16}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
ta có \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=3\)
=> \(VT\le\frac{1}{16\cdot3}=\frac{3}{16}\)
dấu bằng xảy ra khi x=y=z=1
a.\(\frac{1-3x}{2}-\frac{x+3}{2}=\frac{1-3x-x-3}{2}=\frac{1-4x-3}{2}=\frac{-4x-2}{2}=\frac{-2\left(2x+1\right)}{2}=-2x-1\)
b. \(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}=\frac{2\left(x^2-y^2\right)+2y^2}{x}=\frac{2x^2-2y^2+2y^2}{x}=2x\)
c. \(\frac{3x+1}{x+y}-\frac{2x-3}{x+y}=\frac{3x+1-2x+3}{x+y}=\frac{x+4}{x+y}\)
d. \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}=\frac{xy}{2x-y}-\frac{1-x^2}{2x-y}=\frac{xy-1+x^2}{2x-y}\)
e. \(\frac{4x-1}{3x^2y}-\frac{7x-1}{3x^2y}=\frac{4x-1-7x+1}{3x^2y}=\frac{-3x}{3x^2y}=\frac{-1}{xy}\)
a) Rút gọn :
Ta có : \(A=\frac{y-x}{xy}:\left[\frac{y^2}{\left(x-y\right)^2}-\frac{2x^2y}{\left(x^2-y^2\right)^2}+\frac{x^2}{y^2-x^2}\right]\)
\(=\frac{y-x}{xy}:\left[\frac{y^2\left(x+y\right)^2-2x^2y-x^2\left(x^2-y^2\right)}{\left(x^2-y^2\right)^2}\right]\)
\(=\frac{y-x}{xy}:\left[\frac{y^2\left(x^2+2xy+y^2\right)-2x^2y-x^4+x^2y^2}{\left(x^2-y^2\right)^2}\right]\)
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