cho x và y>0 và x+y=xy tìm min S=x+y
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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
Nguyễn Việt Lâm
Giáo viên
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:
\(S+\dfrac{x+y+32}{100}+\dfrac{2}{5}\ge\dfrac{3\left(x+y\right)}{20}+\dfrac{2021}{2022}\)
\(S\ge\dfrac{9}{20}\left(x+y\right)-\dfrac{42}{25}+\dfrac{2021}{2022}\ge\dfrac{9}{20}.2\sqrt{xy}-\dfrac{42}{25}+\dfrac{2021}{2022}=...\)
HN
1
19 tháng 4 2023
\(A=\dfrac{\left(x-y\right)^2+2xy}{x-y}=x-y+\dfrac{2xy}{x-y}=x-y+\dfrac{2}{x-y}>=2\sqrt{2}\)
Dấu = xảy ra khi \(\left\{{}\begin{matrix}x=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\y=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)
T
0
LV
0
DS
1
10 tháng 6 2019
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 \(x=2+\sqrt{2}\),\(y=2-\sqrt{2}\)
LP
1
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: \(x=y=2\)