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Cho x,y > 0. Tìm GTNN của:
a) x2 + y2 + \(\dfrac{1}{xy}\) với x + y = 2
b) x + y + \(\dfrac{1}{xy}\)
a ) Áp dụng BĐT Cô-si với 2 số x ; y > 0 , ta có :
\(x^2+y^2+\dfrac{1}{xy}\ge\dfrac{\left(x+y\right)^2}{2}+\dfrac{1}{\dfrac{\left(x+y\right)^2}{4}}=\dfrac{2^2}{2}+\dfrac{1}{\dfrac{2^2}{4}}=2+1=3\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=1\)
Vậy ...
b ) Áp dụng BĐT Cô-si với 2 số x ; y > 0 , ta có :
\(x+y+\dfrac{1}{xy}\ge3\sqrt[3]{xy.\dfrac{1}{xy}}=3\)
Dấu " = " xảy ra \(\Leftrightarrow x=y=\dfrac{1}{xy}\)
\(\Leftrightarrow x^2y=y^2x=1\)
\(\Leftrightarrow x^3y^3=1\Leftrightarrow xy=1\left(x;y>0\right)\)
\(\Leftrightarrow x=y=1\)
Vậy ...
a: \(M=\dfrac{x+6\sqrt{x}-3\sqrt{x}+18-x}{x-36}\)
\(=\dfrac{3\left(\sqrt{x}+6\right)}{x-36}=\dfrac{3}{\sqrt{x}-6}\)
b: \(N=\dfrac{x^2}{y}\cdot\sqrt{xy\cdot\dfrac{y}{x}}-x^2\)
\(=\dfrac{x^2}{y}\cdot y-x^2=0\)
\(P=x^2y^2+1+1+\frac{1}{x^2y^2}=x^2y^2+2+\frac{1}{256x^2y^2}+\frac{255}{256x^2y^2}\)
\(\ge x^2y^2+\frac{1}{256x^2y^2}+2+\frac{255}{256.\left[\frac{\left(x+y\right)^2}{4}\right]^2}\ge2\sqrt{x^2y^2.\frac{1}{256x^2y^2}}+2+\frac{255}{256.\frac{1}{16}}\)
\(=\frac{1}{8}+2+\frac{255}{16}=\frac{289}{16}\)
Dấu "=" xảy ra khi \(x=y=\frac{1}{2}\)
\(a.x^2-2xy+6y^2-12x+2y+41\)
\(=x^2-2xy+y^2-12x+12y+36+5y^2-10y+5\)
\(=\left(x-y\right)^2-2.6\left(x-y\right)+36+5\left(y-1\right)^2\)
\(=\left(x-y-6\right)^2+5\left(y-1\right)^2\) ≥ \(0\)
\(b.\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}-\dfrac{2x}{y}-\dfrac{2y}{x}+3\)
\(=\dfrac{x^2}{y^2}-2.\dfrac{x}{y}+1+\dfrac{y^2}{x^2}-2.\dfrac{y}{x}+1+1\)
\(=\left(\dfrac{x}{y}-1\right)^2+\left(\dfrac{y}{x}-1\right)^2+1>0\)
Ta có : \(A=x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}+2\left(\frac{x}{y}+\frac{y}{x}\right)\)
\(A=4+\frac{x^2+y^2}{x^2y^2}+\frac{2.\left(x^2+y^2\right)}{xy}=4+\frac{4}{x^2y^2}+\frac{8}{xy}\)
\(A=4\left(\frac{1}{xy}+1\right)^2\)
Mặt khác : \(xy\le\frac{x^2+y^2}{2}=2\Rightarrow\frac{1}{xy}\ge\frac{1}{2}\)
\(\Rightarrow A\ge4\left(\frac{1}{2}+1\right)^2=9\)
Vậy Min A = 9 khi x = y = \(\sqrt{2}\)
\(1>=\left(x+y\right)^2>=\left(2\sqrt{xy}\right)^2=4xy\Rightarrow1>=4xy\Rightarrow\frac{1}{2}>=2xy\)(bđt cosi)
\(\Rightarrow\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}=\left(\frac{1}{x^2+y^2}+\frac{1}{2xy}\right)+\frac{1}{2xy}>=\frac{4}{x^2+2xy+y^2}+\frac{1}{\frac{1}{2}}\)
\(=\frac{4}{\left(x+y\right)^2}+2>=\frac{4}{1^2}+2=4+2=6\)
dấu = xảy ra khi \(x=y=\frac{1}{2}\)
vậy min \(\frac{1}{x^2+y^2}+\frac{1}{xy}=6\)khi \(x=y=\frac{1}{2}\)
a) \(4xy\le\left(x+y\right)^2=1\)
=> \(xy\le4\)
Dấu "=" xảy ra <=> x = y = 1/2
b) A = \(A=x^2+2+\dfrac{1}{x^2}+y^2+2+\dfrac{1}{y^2}=\left(x^2+y^2\right)+\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+4\ge2xy+\dfrac{2}{xy}+4=\left(32xy+\dfrac{2}{xy}\right)-30xy+4\ge8-\dfrac{30}{4}+4=\dfrac{9}{2}\)
Dấu "=" xảy ra <=> x = y = 1/2
a)x2+y2=2 =>(x+y)2-2xy=2<=>-2xy=2-(x+y)2 <=> xy=\(-\dfrac{2-\left(x+y\right)2}{2}\)
mà \(-\dfrac{2-\left(x+y\right)2}{2}< 1\)
=>xy <1