Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
`<=>2P=10x+6y+24/x+32/y`
`<=>2P=6x+24/x+2y+32/y+4x+4y`
`<=>2P=6(x+4/x)+2(y+16/y)+4(x+y)`
Áp dụng BĐT cosi:
`x+4/x>=4=>6(x+4/x)>=24`
`y+16/y>=8=>2(y+16/y)>=16`
Mà `x+y>=6=>4(x+y)>=24`
`=>2P>=24+16+24=64`
`=>P>=32`
Dấu "=" `<=>x=2,y=4`
\(B=x+y+\dfrac{6}{x}+\dfrac{24}{y}=\left(\dfrac{3x}{2}+\dfrac{6}{x}\right)+\left(\dfrac{3y}{2}+\dfrac{24}{y}\right)-\dfrac{3}{2}\left(x+y\right)\)
\(B\ge2\sqrt{\dfrac{18x}{2x}}+2\sqrt{\dfrac{72y}{2y}}-\dfrac{3}{2}.6=15\)
\(B_{min}=15\) khi \(\left(x;y\right)=\left(2;4\right)\)
Ta có \(P^2=\left(\sum\dfrac{x}{\sqrt{y}}\right)^2=\sum\dfrac{x^2}{y}+2\left(\sum\dfrac{xy}{\sqrt{yz}}\right)\)
Mà \(\dfrac{x^2}{y}+\dfrac{xy}{\sqrt{yz}}+\dfrac{xy}{\sqrt{yz}}+z\ge4\sqrt[4]{x^4}=4x\)
Tương tự rồi cộng lại, ta có
\(P^2+x+y+z\ge4\left(x+y+z\right)\Rightarrow P^2\ge3\left(x+y+z\right)=36\Rightarrow P\ge6\)
Ta có: \(P+\frac{1}{2}(a+b)=(\frac{3}{2}x+\frac{6}{x})+(\frac{3}{2}y+\frac{24}{y})\geq 2.3+2.6=18\)
Mà \(a+b\leq 6\) suy ra \(P\geq 15\)
dấu = xảy ra \(<=> x+y=6 , \frac{3}{2}x=\frac{6}{x}\) và \(\frac{3}{2}y=\frac{24}{y}\)
\(<=> x=2 , y=4\)
Đặt A = ( \(\dfrac{3x}{2}\) + \(\dfrac{6}{x}\) ) + ( \(\dfrac{3y}{2}\) + \(\dfrac{24}{y}\) ) - ( \(\dfrac{x+y}{2}\) )
Áp dụng BĐT Cô-si ta có
\(\dfrac{3x}{2}+\dfrac{6}{x}\ge6\)
\(\dfrac{3y}{2}+\dfrac{24}{y}\ge6\)
Có x + y \(\le6\)
=> - (x + y) \(\ge6\) => \(\dfrac{-\left(x+y\right)}{2}\ge3\)
=> A \(\ge15\)
Dấu " = " xảy ra <=> x = 2; y = 4
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
Đặt \(\dfrac{x-y}{z}=m,\dfrac{y-z}{x}=n,\dfrac{z-x}{y}=p\), ta có:
\(\left(m+n+p\right)\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)=3+\dfrac{n+p}{m}+\dfrac{p+m}{n}+\dfrac{m+n}{p}\)
Tính \(\dfrac{n+p}{m}\) theo x, y, z ta được:
\(\dfrac{n+p}{m}=\dfrac{z}{x-y}.\dfrac{y^2-yz+xz-x^2}{xy}=\dfrac{z}{xy}\left(-x-y+x\right)\)
\(=\dfrac{z}{xy}\left(-x-y-z+2z\right)=\dfrac{2x^2}{xy}\) vì \(\left(x+y+z\right)=0\)
Tương tự: \(\dfrac{m+p}{n}=\dfrac{2x^2}{yz}.\dfrac{m+n}{p}=\dfrac{2y^2}{xz}\)
Vậy \(\left(m+n+p\right)\left(\dfrac{1}{m}+\dfrac{1}{n}+\dfrac{1}{p}\right)=3+\dfrac{2\left(x^3+y^3+z^3\right)}{xyz}=3+\dfrac{2.3xyz}{xyz}=3+6=9\)
Đặt \(P=\left(\dfrac{x-y}{z}+\dfrac{y-z}{x}+\dfrac{z-x}{y}\right)\left(\dfrac{z}{x-y}+\dfrac{x}{y-z}+\dfrac{y}{z-x}\right)=9\)
Đặt \(\left\{{}\begin{matrix}\dfrac{x-y}{z}=a\\\dfrac{y-z}{x}=b\\\dfrac{x-z}{y}=c\end{matrix}\right.\)
\(\Leftrightarrow P=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\\ =1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\\ =3+\dfrac{a+c}{b}+\dfrac{a+b}{c}+\dfrac{b+c}{a}\)
Ta có \(\dfrac{a+c}{b}=\dfrac{\dfrac{x-y}{z}+\dfrac{z-x}{y}}{\dfrac{y-z}{x}}=\dfrac{xy-y^2+z^2-xz}{yz}\cdot\dfrac{x}{y-z}\)
\(=\dfrac{\left(z-y\right)\left(y+z-x\right)x}{yz\left(y-z\right)}=\dfrac{x\left(x-y-z\right)}{yz}\)
Mà \(x+y+z=0\Leftrightarrow x=-y-z\)
\(\Leftrightarrow\dfrac{a+c}{b}=\dfrac{x\left(x+x\right)}{yz}=\dfrac{2x^2}{yz}\)
Cmtt ta được \(\dfrac{a+b}{c}=\dfrac{2y^2}{xz};\dfrac{b+c}{a}=\dfrac{2z^2}{xy}\)
Cộng vế theo vế
\(\Leftrightarrow P=\dfrac{2x^2}{yz}+\dfrac{2y^2}{xz}+\dfrac{2z^2}{xy}+3=\dfrac{2x^3+2y^3+2z^3}{xyz}+3\\ \Leftrightarrow P=\dfrac{2\left(x^3+y^3+z^3\right)}{xyz}+3\)
Lại có \(x+y+z=0\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)
\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\\ \Leftrightarrow x^3+y^3+z^3=3xyz\)
Thế vào \(P\)
\(\Leftrightarrow P=\dfrac{2\cdot3xyz}{xyz}+3=6+3=9\)
\(\Leftrightarrow2P=6x+4y+\dfrac{12}{x}+\dfrac{16}{y}\\ \Leftrightarrow2P=\left(\dfrac{12}{x}+3x\right)+\left(\dfrac{16}{y}+y\right)+3\left(x+y\right)\\ \Leftrightarrow2P\ge2\sqrt{36}+2\sqrt{16}+3\cdot6=12+8+18=38\\ \Leftrightarrow P\ge19\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}3x^2=12\\y^2=16\\x+y=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)
\(P=\dfrac{6}{x}+\dfrac{3}{2}x+\dfrac{24}{y}+\dfrac{3}{2}y-\dfrac{1}{2}\left(x+y\right)\ge2\sqrt{6.\dfrac{3}{2}}+2\sqrt{24.\dfrac{3}{2}}-\dfrac{1}{2}.6=15\Rightarrow min=15\Leftrightarrow x=2;y=4\)