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\(\left(\sqrt{x}+\sqrt{y}\right)\left(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right).\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{xy}}\)
\(=\frac{-y+\sqrt{x}.\sqrt{y}}{\sqrt{y}}.\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\frac{\left(\sqrt{x}.\sqrt{y}-y\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}}\)
\(=\frac{xy-y^2}{y}\)
\(=\frac{y\left(x-y\right)}{y}\)
= x - y (đpcm)
b/ \(a-\frac{1}{a}=\sqrt{a}+\frac{1}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}-\frac{1}{\sqrt{a}}=1\)
\(\Leftrightarrow a+\frac{1}{a}-2=1\)
\(\Leftrightarrow a+\frac{1}{a}=3\)
\(\Leftrightarrow a^2+\frac{1}{a^2}+2=9\)
\(\Leftrightarrow\left(a-\frac{1}{a}\right)^2=5\)
\(\Leftrightarrow a-\frac{1}{a}=\sqrt{5}\)
a/ Ta có: \(x=\frac{1-5y}{2}\) thê vô ta được
\(x^2+y^2=y^2+\left(\frac{1-5y}{2}\right)^2=\frac{29y^2-10y+1}{4}\)
\(=\frac{1}{116}\left(29^2y^2-290y+29\right)=\frac{1}{116}\left[\left(29^2y^2-2.29y.5+25\right)+4\right]\)
\(=\frac{1}{116}\left[\left(29y-5\right)^2+4\right]\ge\frac{4}{116}=\frac{1}{29}\)
Cách khác:
\(\frac{\left(x+y\right)^2}{2}+\frac{\left(x+y\right)}{4}\ge2xy+\frac{x+y}{4}\)
\(=\frac{4xy+x+4xy+y}{4}=\frac{x\left(4y+1\right)+y\left(4x+1\right)}{4}\)
\(\ge\frac{4x\sqrt{y}+4y\sqrt{x}}{4}=x\sqrt{y}+y\sqrt{x}\)
Dấu = xảy ra khi \(x=y=\frac{1}{4}\)
\(\frac{1}{2}\left(x+y\right)\left(x+y+\frac{1}{2}\right)=\frac{1}{2}\left(x+y\right)\left(x+\frac{1}{4}+y+\frac{1}{4}\right)\)
Áp dụng bất đẳng thức cauchy:
\(x+y\ge2\sqrt{xy}\)
\(x+\frac{1}{4}\ge2\sqrt{\frac{x}{4}}=\sqrt{x}\)
\(y+\frac{1}{4}\ge2\sqrt{\frac{y}{4}}=\sqrt{y}\)
do đó \(VT\ge\frac{1}{2}.2.\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)=x\sqrt{y}+y\sqrt{x}\)(đpcm)
Dấu = xảy ra khi \(x=y=\frac{1}{4}\)
\(a,\)\(2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\Rightarrow2a^2+2b^2\ge a^2+2ab+b^2\)
\(\Rightarrow a^2+b^2\ge2ab\)
\(\Rightarrow a^2-2ab+b^2\ge0\)
\(\Rightarrow\left(a-b\right)^2\ge0\) ( luôn đúng )
\(\Rightarrow2\left(a^2+b^2\right)\ge\left(a+b\right)^2\)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1^2}{x}+\frac{1^2}{y}+\frac{1^2}{z}\ge\frac{\left(1+1+1\right)^2}{x+y+z}=\frac{9}{x+y+z}\)( Bất đẳng thức Svac-xơ )
Dấu = xảy ra khi \(\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\)
BĐT trên
\(< =>\frac{xy+yz+xz}{xyz}\ge\frac{9}{x+y+z}\)
\(< =>\left(x+y+z\right)\left(xy+yz+xz\right)\ge9xyz\)
Áp dụng BĐT cô si cho 3 số :
\(x+y+z\ge3\sqrt[3]{xyz}\)
\(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
Nhân vế với vế : \(\left(x+y+z\right)\left(xy+yz+xz\right)\ge3\sqrt[3]{xyz}.3\sqrt[3]{x^2y^2z^2}=9xyz\)
Nên ta có đpcm
a) Áp dụng BĐT AM-GM ta có:
\(x+y\ge2\sqrt{xy}\)
\(\Rightarrow\)\(\frac{x+y}{2}\ge\sqrt{xy}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y\)
b) Áp dụng BĐT AM-GM ta có:
\(\frac{\sqrt{x}}{\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{x}}\ge2\sqrt{\frac{\sqrt{x}}{\sqrt{y}}.\frac{\sqrt{y}}{\sqrt{x}}}=2\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y\)