Cho x,y>0 và \(\sqrt{xy}.\left(x-y\right)=x+y\)Tìm min A=x+y
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
xy(x-y)2=(x+y)2 ĐK:x>y
(x+y)2=[(x+y)2-4xy]xy
(x+y)2(xy-1)=4x2y2
\(\frac{1}{\left(x+y\right)^2}=\frac{xy-1}{4x^2y^2}=\frac{1}{4}\left(\frac{1}{xy}-\frac{1}{x^2y^2}\right)\)
\(\frac{1}{\left(x+y\right)^2}=\left[-\left(\frac{1}{xy}-\frac{1}{2}\right)^2+\frac{1}{4}\right]\le\frac{1}{16}\)
=> \(x+y\ge4\)
Dấu "=" xảy ra khi \(x=2+\sqrt{2}\),\(y=2-\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Thử nhé
Vì P là bất đẳng thức đối xứng nên dự đoán điểm rơi \(x=y=z=\dfrac{\sqrt{2021}}{3}\)
Thay vo P ta duoc \(P=4.\sqrt{2021}\)
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\(P=\sum\dfrac{\left(x+y\right)\sqrt{\left(y+z\right)\left(z+x\right)}}{z}\)
Cauchy-Schwarz:
\(\Rightarrow\left(y+z\right)\left(z+x\right)\ge\left(z+\sqrt{xy}\right)^2\Rightarrow\sqrt{\left(y+z\right)\left(z+x\right)}\ge z+\sqrt{xy}\)
\(\Rightarrow P\ge\sum\dfrac{\left(x+y\right)\left(z+\sqrt{xy}\right)}{z}\ge\sum\dfrac{xz+yz+x\sqrt{y}+y\sqrt{x}}{z}=\sum x+y+\dfrac{\left(x+y\right)\sqrt{xy}}{z}\ge\sum x+y+\dfrac{2xy}{z}\)
\(\Rightarrow P\ge2(x+y+z)+2\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\)
Cauchy-Schwarz: \(\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\left(\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{zx}{y}\right)\ge\left(\sqrt{\dfrac{xy}{z}.\dfrac{yz}{z}}+\sqrt{\dfrac{yz}{x}.\dfrac{zx}{y}}+\sqrt{\dfrac{zx}{y}.\dfrac{xy}{z}}\right)^2=\left(x+y+z\right)^2\)
\(\Rightarrow P\ge2(x+y+z)+2\left(x+y+z\right)=4\left(x+y+z\right)=4\sqrt{2021}\)
\("="\Leftrightarrow x=y=z=\dfrac{\sqrt{2021}}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Do \(x-y=\dfrac{x+y}{\sqrt{xy}}>0\Rightarrow x>y\)
Khi đó:
\(\sqrt{xy}\left(x-y\right)=x+y\Rightarrow xy\left(x-y\right)^2=\left(x+y\right)^2\)
\(\Rightarrow xy\left[\left(x+y\right)^2-4xy\right]=\left(x+y\right)^2\)
\(\Rightarrow\left(xy-1\right)\left(x+y\right)^2=4x^2y^2\)
\(\Rightarrow\left(x+y\right)^2=\dfrac{4x^2y^2}{xy-1}\)
Do vế trái dương nên vế phải dương \(\Rightarrow xy-1>0\)
\(\Rightarrow\left(x+y\right)^2=\dfrac{4x^2y^2-4+4}{xy-1}=4xy+4+\dfrac{4}{xy-1}=4\left(xy-1\right)+\dfrac{4}{xy-1}+8\)
\(\ge2\sqrt{4\left(xy-1\right).\dfrac{4}{xy-1}}+8=16\)
\(\Rightarrow x+y\ge4\)
\(P_{min}=4\) khi \(\left(x;y\right)=\left(2+\sqrt{2};2-\sqrt{2}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\dfrac{x^2+y^2+2xy}{x^2+y^2}+\dfrac{x^2+y^2+2xy}{xy}\)
\(=1+\dfrac{2xy}{x^2+y^2}+2+\dfrac{x^2+y^2}{xy}\)
\(=3+\dfrac{2xy}{x^2+y^2}+\dfrac{x^2+y^2}{2xy}+\dfrac{x^2+y^2}{2xy}\)
\(\dfrac{2xy}{x^2+y^2}+\dfrac{x^2+y^2}{2xy}>=2\cdot\sqrt{\dfrac{2xy}{x^2+y^2}\cdot\dfrac{x^2+y^2}{2xy}}=2\)
Dấu = xảy ra khi \(\dfrac{x^2+y^2}{2xy}=\dfrac{2xy}{x^2+y^2}\)
=>x=y
x^2+y^2>=2xy
=>\(\dfrac{x^2+y^2}{2xy}>=1\)
Dấu = xảy ra khi x=y
=>S>=6
Dấu = xảy ra khi x=y