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Ta có \(xy+yz+xz=\frac{2^2-18}{2}=-7\)
\(x+y+z=2\)=> \(z-1=-x-y+1\)
=> \(\frac{1}{xy+z-1}=\frac{1}{xy-x-y+1}=\frac{1}{\left(x-1\right)\left(y-1\right)}\)
Tương tự \(\frac{1}{yz+x-1}=\frac{1}{\left(y-1\right)\left(z-1\right)};\frac{1}{xz+y-1}=\frac{1}{\left(z-1\right)\left(x-1\right)}\)
=> \(S=\frac{x+y+z-3}{\left(x-1\right)\left(y-1\right)\left(z-1\right)}=-\frac{1}{xyz-\left(yz+xy+xz\right)+\left(x+y+z\right)-1}\)
\(=\frac{-1}{-1+7+2-1}=-\frac{1}{7}\)
Vậy \(S=-\frac{1}{7}\)
Từ \(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{xz}{x+z}\Rightarrow\frac{x+y}{xy}=\frac{y+z}{yz}=\frac{x+z}{xz}\)
\(\Rightarrow\frac{x}{xy}+\frac{y}{xy}=\frac{y}{yz}+\frac{z}{yz}=\frac{x}{xz}+\frac{z}{xz}\)
\(\Rightarrow\frac{1}{y}+\frac{1}{x}=\frac{1}{y}+\frac{1}{z}=\frac{1}{z}+\frac{1}{x}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\Rightarrow x=y=z\).Khi đó
\(P=\frac{20xy+4yz+2013xz}{x^2+y^2+z^2}=\frac{20x^2+4x^2+2013x^2}{x^2+x^2+x^2}=\frac{2037x^2}{3x^2}=679\)
cho x,y>0 thỏa mãn \(^{x^2+y^2-xy=8}\)
tìm GTNN và GTNN của biểu thức M=\(^{x^2+y^2}\)
Ta luôn có:
\(xy+yz+zx\le x^2+y^2+z^2\)\(=3\); dấu "=" xảy ra ⇔\(x=y=z\)
\(x\le\frac{x^2+1}{2}\); dấu "=" xảy ra ⇔ \(x=1\)
\(y\le\frac{y^2+1}{2}\); dấu "=" xảy ra ⇔ \(y=1\)
\(z\le\frac{z^2+1}{2}\); dấu "=" xảy ra ⇔ \(z=1\)
Suy ra: \(x+y+z\le\frac{x^2+y^2+z^2+3}{2}=\frac{6}{2}=3\)
Do đó: \(P_{max}=xy+yz+zx+\frac{5}{x+y+z}\le3+\frac{5}{3}=\frac{14}{3}\)
Dấu "=" xảy ra ⇔ x=y=z=1
a)Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
\(\Rightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2=3^2=9\)
\(\Rightarrow3A\ge9\Rightarrow A\ge3\)
Đẳng thức xảy ra khi \(x=y=z=1\)
b)Ta có BĐT \(3\left(ab+bc+ca\right)\le\left(a+b+c\right)^2\Leftrightarrow-\left(a+b+c\right)^2\le0\)
\(\Rightarrow3B\le\left(x+y+z\right)^2=3^2=9\)
\(\Rightarrow B\le3\)
Đẳng thức xảy ra khi \(x=y=z=1\)
\(\frac{3}{2}x^2+y^2+z^2+yz=1\Leftrightarrow3x^2+2y^2+2z^2+2yz=2\)
\(\Leftrightarrow\left(x^2+y^2+z^2+2xy+2yz+2zx\right)+\left(x^2-2xy+y^2\right)+\left(x^2-2xz+z^2\right)=2\)
\(\Leftrightarrow\left(x+y+z\right)^2+\left(x-y\right)^2+\left(x-z\right)^2=2\)
Suy ra : \(A^2\le2\Rightarrow A\le\sqrt{2}\)
Vậy Max A = \(\sqrt{2}\) khi \(\hept{\begin{cases}x=y\\x=z\\x+y+z=\sqrt{2}\end{cases}\Leftrightarrow}x=y=z=\frac{\sqrt{2}}{3}\)
Ta có : \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)
\(\Leftrightarrow x^2+y^2+z^2+2.\left(xy+yz+zx\right)\ge xy+yz+zx+2.\left(xy+yz+zx\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2\ge3.\left(xy+yz+zx\right)\)
\(\Leftrightarrow xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}=\frac{3^2}{3}=2\)
Hay : \(B\le3\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Vậy \(GTLN\) của \(B=3\) khi \(x=y=z=1\)
Ta có bất đẳng thức sau : \(xy+yz+zx\le x^2+y^2+z^2\)
\(< =>2\left(xy+yz+zx\right)\le2\left(x^2+y^2+z^2\right)\)
\(< =>2xy+2yz+2zx\le2x^2+2y^2+2z^2\)
\(< =>2x^2+2y^2+2z^2-2xy-2yz-2zx\ge0\)
\(< =>\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2zx+x^2\right)\ge0\)
\(< =>\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)*đúng*
Khi đó ta được bất đăng thức \(xy+yz+zx\le x^2+y^2+z^2\)
\(< =>3\left(xy+yz+zx\right)\le\left(x+y+z\right)^2=3^2=9\)
\(< =>xy+yz+zx\le\frac{9}{3}=3\) Tương đương \(B\le3\)
Dấu "=" xảy ra khi và chỉ khi \(x=y=z=1\)
Vậy GTLN của B = 3 đạt được khi x = y = z = 1
Giải:
Ta có:
\(P=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\)
\(\Leftrightarrow P=\dfrac{1}{2}\left[\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)+\left(\dfrac{yz}{x}+\dfrac{zx}{y}\right)+\left(\dfrac{zx}{y}+\dfrac{xy}{z}\right)\right]\)
Áp dụng BĐT AM-GM, có:
\(P=\dfrac{1}{2}\left[\left(\dfrac{xy}{z}+\dfrac{yz}{x}\right)+\left(\dfrac{yz}{x}+\dfrac{zx}{y}\right)+\left(\dfrac{zx}{y}+\dfrac{xy}{z}\right)\right]\ge\dfrac{1}{2}.\left(2\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}+2\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+2\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)\)
\(\Leftrightarrow P\ge\sqrt{\dfrac{xy}{z}.\dfrac{yz}{x}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\)
\(\Leftrightarrow P\ge x+y+z\)
\(\Leftrightarrow P\ge2019\)
\(\Leftrightarrow P_{Min}=2019\)
\("="\Leftrightarrow x=y=z=\dfrac{2019}{3}\)
Vậy ...
Theo bđt Bunhiacopxki ta có : \(\left(x^2+y^2+z^2\right)\left(z^2+x^2+y^2\right)\ge\left(xy+yz+xz\right)^2\)
\(\Rightarrow x^2+y^2+z^2\ge\left|xy+yz+xz\right|\ge xy+yz+xz\)
\(\Leftrightarrow x^2+y^2+z^2+2xy+2yz+2xz\ge3\left(xy+yz+xz\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+xz\right)\)
\(\Rightarrow xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{3^2}{3}=3\) có GTLN là 3
Dấu "=" xảy ra khi \(x=y=z=1\)