Bài 5.
\(P=\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\)
Áp dụng BĐT AM-GM, ta có:
\(\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge2\sqrt{\left(2x+\dfrac{1}{x}\right)^2.\left(2y+\dfrac{1}{y}\right)^2}\)
\(\Leftrightarrow\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge2\left(2x+\dfrac{1}{x}\right).\left(2y+\dfrac{1}{y}\right)\)
\(\Leftrightarrow\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge2\left(4xy+\dfrac{2x}{y}+\dfrac{2y}{x}+\dfrac{1}{xy}\right)\)
\(\Leftrightarrow\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge8xy+4\left(\dfrac{x}{y}+\dfrac{y}{x}\right)+\dfrac{2}{xy}\)
Áp dụng BĐT AM-GM, ta có:
\(\dfrac{x}{y}+\dfrac{y}{x}\ge2\sqrt{\dfrac{xy}{xy}}=2\sqrt{1}=2\)
\(\Leftrightarrow4\left(\dfrac{x}{y}+\dfrac{y}{x}\right)\ge8\)
\(8xy+\dfrac{2}{xy}\ge2\sqrt{\dfrac{8xy.2}{xy}}=2\sqrt{16}=8\)
\(\Rightarrow\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge16\)
Dấu "=" xảy ra khi:
\(\dfrac{x}{y}=\dfrac{y}{x}\)
\(\Leftrightarrow x^2=y^2\)
\(\Leftrightarrow x=y=\dfrac{1}{2}\)
Vậy \(MinP=16\) khi \(x=y=\dfrac{1}{2}\)
a) -△BDE và △DCE có: \(\widehat{BDE}=\widehat{DCE}=90^0;\widehat{E}\) góc chung.
\(\Rightarrow\)△BDE∼△DCE (g-g).
b) -△CHD và △DCB có: \(\widehat{CHD}=\widehat{DCB}=90^0;\widehat{HCD}=\widehat{CDB}\)
\(\Rightarrow\)△CHD∼△DCB (g-g) \(\Rightarrow\dfrac{CD}{CH}=\dfrac{DB}{DC}\Rightarrow CD^2=DB.CH\)
\(CD=AB=6\left(cm\right)\)
-△BCD vuông tại C \(\Rightarrow BD^2=BC^2+DC^2\Rightarrow BD=\sqrt{BC^2+DC^2}=\sqrt{6^2+8^2}=10\left(cm\right)\)
\(CD^2=DB.CH\Rightarrow CH=\dfrac{CD^2}{DB}=\dfrac{8^2}{10}=6,4\left(cm\right)\)
c) ABCD là HCN mà AC cắt BD tại O \(\Rightarrow\)O là trung điểm BD.
-△ODE có: HK//OD \(\Rightarrow\dfrac{HK}{OD}=\dfrac{EK}{EO}\)
-△OBE có: CK//OB \(\Rightarrow\dfrac{CK}{OB}=\dfrac{EK}{EO}=\dfrac{HK}{OD}\)
\(\Rightarrow CK=BK\) nên K là trung điểm BC.
d) -△BDE có: \(\dfrac{HE}{HD}.\dfrac{OD}{OB}.\dfrac{CB}{CE}=1\) và H thuộc đoạn DE, O thuộc đoạn BD, C thuộc đoạn BE.
\(\Rightarrow\)OE, BH, DC đồng quy (định lí Ceva đảo).
-Áp dụng BĐT AM-GM ta có:
\(P=\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2\ge2.\left(2x+\dfrac{1}{x}\right).\left(2y+\dfrac{1}{y}\right)\ge2.\sqrt{2x}.\sqrt{\dfrac{1}{x}}.\sqrt{2y}.\sqrt{\dfrac{1}{y}}=2.\sqrt{\dfrac{2x}{x}}.\sqrt{\dfrac{2y}{y}}=2.2.2=16\)
\(P_{min}=16\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{1}{x}=2y+\dfrac{1}{y}\\2x=\dfrac{1}{x}\\2y=\dfrac{1}{y}\end{matrix}\right.\Leftrightarrow x=y=\dfrac{1}{\sqrt{2}}\)hay \(x=y=-\dfrac{1}{\sqrt{2}}\)
\(x,y>0\)
\(x+y=1\Leftrightarrow\left(x+y\right)^2=1\le2\left(x^2+y^2\right)\Leftrightarrow x^2+y^2\ge\dfrac{1}{2}\)
\(x+y=1\Leftrightarrow\left(x+y\right)^2=1\ge4xy\Leftrightarrow xy\le\dfrac{1}{4}\Rightarrow\dfrac{1}{xy}\ge4\Rightarrow\dfrac{2}{xy}\ge8\)
\(P=\left(2x+\dfrac{1}{x}\right)^2+\left(2y+\dfrac{1}{y}\right)^2=4x^2+2.2x.\dfrac{1}{x}+\dfrac{1}{x^2}+4y^2+2.2y.\dfrac{1}{y}+\dfrac{1}{y^2}=4\left(x^2+y^2\right)+\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+8\ge4.\dfrac{1}{2}+2.\dfrac{1}{xy}+4\ge4.\dfrac{1}{2}+8+8=18\)\(P_{min}=18\) \(\Leftrightarrow\left[{}\begin{matrix}x=y\\x^2+y^2=\dfrac{1}{2}\\\dfrac{2}{xy}=8\end{matrix}\right.\Leftrightarrow x=y=\dfrac{1}{2}\)
