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15 tháng 6 2021

Ta có: \(x+y=xy\Leftrightarrow\frac{1}{x}+\frac{1}{y}=1\)

Mà \(\frac{1}{x}+\frac{1}{y}\ge\frac{\left(1+1\right)^2}{x+y}=\frac{4}{x+y}\)

\(\Rightarrow\frac{4}{x+y}\le1\Rightarrow x+y\ge4\)

Dấu "=" xảy ra khi: \(x=y=2\)

15 tháng 6 2021

Áp dụng bất đẳng thức cosi ta có:

`x+y>=2\sqrt{xy}`

Mà `x+y=xy`

`=>xy>=2\sqrt{xy}`

`x,y>0=>xy>0` chia hai vế cho `2sqrt{xy}>0` ta có:

`\sqrt{xy}>=2`

`<=>xy>=4`

`=>S>=4`

Dấu "=" xảy ra khi `x=y=2`

20 tháng 7 2019

\(1,A=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)

                                             \(\ge\frac{4}{\left(x+y^2\right)}+\frac{1}{\frac{\left(x+y\right)^2}{2}}\ge\frac{4}{1}+\frac{2}{1}=6\)

Dấu "=" <=> x= y = 1/2

20 tháng 7 2019

\(2,A=\frac{x^2+y^2}{xy}=\frac{x}{y}+\frac{y}{x}=\left(\frac{x}{9y}+\frac{y}{x}\right)+\frac{8x}{9y}\ge2\sqrt{\frac{x}{9y}.\frac{y}{x}}+\frac{8.3y}{9y}\)

                                                                                                  \(=2\sqrt{\frac{1}{9}}+\frac{8.3}{9}=\frac{10}{3}\)

Dấu "=" <=> x = 3y

29 tháng 3 2022

\(S=\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y^3}{16\left(x+16\right)}+\dfrac{2021}{2022}\)

\(\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y+16}{100}+\dfrac{16}{80}\ge3\sqrt[3]{\dfrac{x^3\left(y+16\right).16}{16\left(y+16\right).100.80}}=\dfrac{3x}{20}\)

\(tương\) \(tự\Rightarrow\dfrac{y^3}{16\left(x+16\right)}\ge\dfrac{3y}{20}\)

\(\Rightarrow S\ge\dfrac{3x}{20}+\dfrac{3y}{20}-\left(\dfrac{x+16}{100}+\dfrac{y+16}{100}\right)-2.\dfrac{16}{80}+\dfrac{2021}{2022}=\dfrac{3x+3y}{20}-\dfrac{x+y+32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{15x+15y-x-y-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{14\left(x+y\right)-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}\)

\(xy=16\le\dfrac{\left(x+y\right)^2}{4}\Rightarrow x+y\ge8\Rightarrow S\ge\dfrac{14.8-32}{100}-\dfrac{2}{5}+\dfrac{2021}{2022}=\dfrac{2}{5}+\dfrac{2021}{2022}\)

\(\Rightarrow minS=\dfrac{2}{5}+\dfrac{2021}{2022}\Leftrightarrow x=y=4\)

NV
29 tháng 3 2022

\(\dfrac{x^3}{16\left(y+16\right)}+\dfrac{y+16}{100}+\dfrac{1}{5}\ge3\sqrt[3]{\dfrac{x^3\left(y+16\right)}{16.100.5\left(y+16\right)}}=\dfrac{3x}{20}\)

Tương tự: \(\dfrac{y^3}{16\left(x+16\right)}+\dfrac{x+16}{100}+\dfrac{1}{5}\ge\dfrac{3y}{20}\)

Cộng vế:

\(S+\dfrac{x+y+32}{100}+\dfrac{2}{5}\ge\dfrac{3\left(x+y\right)}{20}+\dfrac{2021}{2022}\)

\(S\ge\dfrac{9}{20}\left(x+y\right)-\dfrac{42}{25}+\dfrac{2021}{2022}\ge\dfrac{9}{20}.2\sqrt{xy}-\dfrac{42}{25}+\dfrac{2021}{2022}=...\)

\(A=\dfrac{\left(x-y\right)^2+2xy}{x-y}=x-y+\dfrac{2xy}{x-y}=x-y+\dfrac{2}{x-y}>=2\sqrt{2}\)

Dấu = xảy ra khi \(\left\{{}\begin{matrix}x=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\y=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)

xy(x-y)2=(x+y)2       ĐK:x>y

(x+y)2=[(x+y)2-4xy]xy

 (x+y)2(xy-1)=4x2y2

\(\frac{1}{\left(x+y\right)^2}=\frac{xy-1}{4x^2y^2}=\frac{1}{4}\left(\frac{1}{xy}-\frac{1}{x^2y^2}\right)\)

\(\frac{1}{\left(x+y\right)^2}=\left[-\left(\frac{1}{xy}-\frac{1}{2}\right)^2+\frac{1}{4}\right]\le\frac{1}{16}\)

=> \(x+y\ge4\)

Dấu "=" xảy ra khi \(x=2+\sqrt{2}\),\(y=2-\sqrt{2}\)

8 tháng 7 2020

\(x+y=\left(x+2\right)+\left(y+2\right)-4\ge2\sqrt{\left(x+2\right)\left(y+2\right)}-4=6\)

12 tháng 7 2020

ta có: S = x+y

=> S=( x+2)+(y+2) - 4

AD BDDT cô-si ta có: \(\left(x+2\right)+\left(y+2\right)\ge2\sqrt{\left(x+2\right).\left(y+2\right)}=2.3=6\)

=> \(S\ge2\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+2=y+2\\\left(x+2\right).\left(y+2\right)=9\end{cases}\Leftrightarrow x=y=1}\)( TM đk x>0; y>0)

KL: MinS = 2 tại x=y=1