可用柯西不等式的基本不等式訓(xùn)練題(含詳解)

可用柯西不等式的基本不等式訓(xùn)練題(含詳解)剔除下面文章的格式錯(cuò)誤，刪除明顯有問(wèn)題的段落,然后再小幅度的改寫(xiě)每段話(huà)。可用柯西不等式的基本不等式訓(xùn)練題（含詳解）1.已知$a>0,b>0,a+b=2$，求$ab$的最大值。解：由柯西不等式得：$(a+b)^2\leq2(a^2+b^2)$，即$2\leq2(a^2+b^2)$，故$a^2+b^2\geq1$，再由$(a-b)^2\geq0$，得$a^2+b^2\geq2ab$，即$ab\leq1$，當(dāng)$a=b=1$時(shí)，等號(hào)成立，故$ab$的最大值為$1$。2.若直線(xiàn)$x+y=8$的最小值是多少？解：由柯西不等式得：$(1^2+1^2)(x^2+y^2)\geq(x+y)^2$，即$2(x^2+y^2)\geq64$，故$x^2+y^2\geq32$，即直線(xiàn)$x+y=8$到原點(diǎn)的距離的平方不小于$32$，故直線(xiàn)$x+y=8$的最小值為$4\sqrt{2}$。3.已知直線(xiàn)$kx-y+2k-1=0$恒過(guò)定點(diǎn)$A$，點(diǎn)$A$也在直線(xiàn)$mx+ny+1=0$上，其中$m,n$均為正數(shù)，則中$m,n$的最小值是多少？解：由題意可得$kx-y+2k-1=0$與$mx+ny+1=0$的交點(diǎn)為點(diǎn)$A$，設(shè)交點(diǎn)坐標(biāo)為$(x_0,y_0)$，則有$\begin{cases}kx_0-y_0+2k-1=0\\mx_0+ny_0+1=0\end{cases}$，解得$x_0=\dfrac{2k-n}{k^2+mn}$，$y_0=\dfrac{nk-2m}{k^2+mn}$，則$A$點(diǎn)坐標(biāo)為$\left(\dfrac{2k-n}{k^2+mn},\dfrac{nk-2m}{k^2+mn}\right)$，由于$A$點(diǎn)在直線(xiàn)$kx-y+2k-1=0$上，故有$\dfrac{nk-2m}{k^2+mn}=2k-1+\dfrac{n-2k^2}{k^2+mn}$，整理得$\dfrac{(m-n)k^2}{k^2+mn}=2k-1$，即$k=\dfrac{mn}{m+n-2}$，代入$kx-y+2k-1=0$中可得直線(xiàn)方程為$\dfrac{mn}{m+n-2}x-y+\dfrac{2mn}{m+n-2}-1=0$，即$mnx-(m+n-2)y+2mn-m^2-n^2+2mn-2m-2n+2=0$，由于$A$點(diǎn)在該直線(xiàn)上，故有$\dfrac{mn}{m+n-2}\cdot\dfrac{2k-n}{k^2+mn}-\dfrac{nk-2m}{k^2+mn}(m+n-2)+2mn-m^2-n^2+2mn-2m-2n+2=0$，整理得$k^2+mn=\dfrac{2mn(m+n)}{m+n-2}$，即$k=\dfrac{mn}{m+n-2}$，故$m+n\geq2\sqrt{mn}$，即$\sqrt{m}+\sqrt{n}\geq2$，故$m+n$的最小值為$4$。4.已知正數(shù)$x,y$滿(mǎn)足$\dfrac{x^2}{3}+y^2=1$，求$x+2y$的最大值。解：由柯西不等式得：$(\dfrac{1}{3}+4)(x^2+y^2)\geq(x+2y)^2$，即$\dfrac{13}{3}\geq(x+2y)^2$，故$x+2y\leq\sqrt{\dfrac{13}{3}}$，當(dāng)$x=\sqrt{3},y=\dfrac{1}{\sqrt{3}}$時(shí)，等號(hào)成立，故$x+2y$的最大值為$\sqrt{\dfrac{13}{3}}$。5.如圖，在$\triangleABC$中，$BD=\dfrac{2}{3}BC$，$E$為線(xiàn)段$AD$上的動(dòng)點(diǎn)，且$CE=xA+yCB$，求$\dfrac{1}{3}BE^2+\dfrac{2}{3}DE^2$的最小值。解：由柯西不等式得：$(x^2+y^2)(CE^2+CB^2)\geq(xCE+yCB)^2$，即$(x^2+y^2)(xE+y)^2\geq(2x^2+2xy+1)^2$，令$t=xE+y$，則$t^2\geq(2x^2+2xy+1)^2/(x^2+y^2)$，即$t^2\geq\dfrac{4x^2+4xy+1}{x^2+y^2}+\dfrac{4x^2y^2}{(x^2+y^2)^2}$，又有$DE=\dfrac{1}{3}t$，$BE=BC-CE=BC-xE-yCB=\dfrac{1}{3}(3-2x)BC$，故$\dfrac{1}{3}BE^2+\dfrac{2}{3}DE^2=\dfrac{4}{9}BC^2(x^2-2x+1)+\dfrac{4}{27}t^2=\dfrac{4}{9}BC^2(x-1)^2+\dfrac{4}{27}t^2+\dfrac{8}{27}BC^2(x-1)$，令$f(x)=\dfrac{4}{9}BC^2(x-1)^2+\dfrac{4}{27}t^2+\dfrac{8}{27}BC^2(x-1)$，則$f'(x)=\dfrac{8}{9}BC^2(x-1)+\dfrac{16}{27}t$，令$f'(x)=0$，解得$x=\dfrac{1}{2}$，代入$f(x)$中得到$f(\dfrac{1}{2})=\dfrac{1}{9}BC^2$，故$\dfrac{1}{3}BE^2+\dfrac{2}{3}DE^2$的最小值為$\dfrac{1}{9}BC^2$。6.若對(duì)$x>0,y>0$，有$\dfrac{x+2y}{\sqrt{xy}}\geq\sqrt{21}$恒成立，則實(shí)數(shù)$m$的取值范圍是多少？解：將不等式化為$\dfrac{x}{\sqrt{21}\sqrt{xy}}+\dfrac{2y}{\sqrt{21}\sqrt{xy}}\geq1$，再由柯西不等式得：$\dfrac{x}{\sqrt{21}\sqrt{xy}}+\dfrac{2y}{\sqrt{21}\sqrt{xy}}\leq\sqrt{\dfrac{x^2}{21xy}+\dfrac{4y^2}{21xy}}=\dfrac{\sqrt{5}}{\sqrt{xy}}$，故$\dfrac{\sqrt{5}}{\sqrt{xy}}\geq1$，即$xy\geq\dfrac{5}{4}$，又由$\dfrac{x+2y}{\sqrt{xy}}\geq\sqrt{21}$，得$x+2y\geq\sqrt{21xy}\geq\sqrt{\dfrac{105}{4}}=\dfrac{\sqrt{21}}{2}$，故$m\leq\dfrac{\sqrt{21}}{2}$。7.圓$x^2+y^2+2x-6y+1=0$關(guān)于直線(xiàn)$ax-by+3=0(a>0,b>0)$對(duì)稱(chēng)，求最小的$a+b$。解：設(shè)圓心為$O(-1,3)$，則直線(xiàn)$ax-by+3=0$的斜率為$k=\dfrac{a}$，過(guò)點(diǎn)$O$且垂直于直線(xiàn)$ax-by+3=0$的直線(xiàn)方程為$y-3=-\dfrac{a}(x+1)$，兩直線(xiàn)的交點(diǎn)為$A$，則$OA$的斜率為$\dfrac{a}$，$A$點(diǎn)坐標(biāo)為$(\dfrac{3a}{a^2+b^2}-1,\dfrac{3b}{a^2+b^2}+3)$，設(shè)對(duì)稱(chēng)點(diǎn)為$B$，則$B$點(diǎn)坐標(biāo)為$(\dfrac{3a}{a^2+b^2}+1,\dfrac{3b}{a^2+b^2}-3)$，由于圓關(guān)于直線(xiàn)$ax-by+3=0$對(duì)稱(chēng)，故$B$點(diǎn)也在圓上，即$(\dfrac{3a}{a^2+b^2}+1)^2+(\dfrac{3b}{a^2+b^2}-3)^2+2(\dfrac{3a}{a^2+b^2}+1)-6(\dfrac{3b}{a^2+b^2}-3)+1=(\dfrac{3a}{a^2+b^2}-1)^2+(\dfrac{3b}{a^2+b^2}+3)^2+2(\dfrac{3a}{a^2+b^2}-1)-6(\dfrac{3b}{a^2+b^2}+3)+1$，整理得$\dfrac{a^2+b^2}{3a-b}=10$，故$a+b=\df