cho x,y >0 , x+y =1. tìm Min
A=\(\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}\)
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B
3
3 tháng 7 2021
Áp dụng BĐT Cauchy-Schwarz dạng Engel có:
\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+y^2+2xy}+\dfrac{1}{\dfrac{\left(x+y\right)^2}{2}}=\dfrac{4}{\left(x+y\right)^2}+\dfrac{2}{\left(x+y\right)^2}=6\)
Dấu "=" xảy ra khi x=y=\(\dfrac{1}{2}\)
3 tháng 7 2021
áp dụng BDT AM-GM
\(=>x+y\ge2\sqrt{xy}=>1\ge2\sqrt{xy}=>\sqrt{xy}\le\dfrac{1}{2}=>xy\le\dfrac{1}{4}\)
\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{1}{2xy}\)
\(\ge\dfrac{4}{x^2+2xy+y^2}+\dfrac{1}{2.\dfrac{1}{4}}=\dfrac{4}{\left(x+y\right)^2}+2=4+2=6\)
dấu"=" xảy ra \(< =>x=y=\dfrac{1}{2}\)
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 \(\left\{{}\begin{matrix}x=\dfrac{\sqrt{6}+\sqrt{2}}{2}\\y=\dfrac{\sqrt{6}-\sqrt{2}}{2}\end{matrix}\right.\)
MH
26 tháng 4 2023
\(P=\dfrac{1}{x^2+y^2+z^2}+\dfrac{2023}{xy+yz+zx}\)
\(=\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy+yz+zx}+\dfrac{1}{xy+yz+zx}+\dfrac{2021}{xy+yz+zx}\)
\(\ge\dfrac{9}{\left(x+y+z\right)^2}+\dfrac{2021}{\dfrac{\left(x+y+z\right)^2}{3}}\)\(=9+\dfrac{2021}{\dfrac{1}{3}}=6072\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Ta có:
+) \(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}\left(\text{Cô si}\right)\)
+) \(\dfrac{1}{x^2+y^2+z^2}+\dfrac{1}{xy+yz+zx}+\dfrac{1}{xy+yz+zx}\)
\(\ge\dfrac{9}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}=\dfrac{9}{\left(x+y+z\right)^2}\left(\text{Svácxơ}\right)\)
DT
1
31 tháng 5 2017
Ta có :
\(K=\frac{1}{x^2+y^2}+\frac{1}{xy}=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\)(1)
Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}>=\frac{4}{a+b}\)( "=" khi a=b ) , ta có :
\(\frac{1}{x^2+y^2}+\frac{1}{2xy}>=\frac{4}{x^2+2xy+y^2}\)
\(\Rightarrow\frac{1}{x^2+y^2}+\frac{1}{2xy}>=\frac{4}{\left(x+y\right)^2}=\frac{4}{1^2}=4\) (2)
Lại có : \(\left(x-y\right)^2>=0\) ("=" khi x=y )
\(\Leftrightarrow x^2-2xy+y^2>=0\)
\(\Leftrightarrow x^2+y^2>=2xy\)
\(\Leftrightarrow x^2+y^2+2xy>=4xy\)
\(\Leftrightarrow\left(x+y\right)^2>=4xy\)
\(\Leftrightarrow1>=4xy\)
\(\Leftrightarrow2xy< =\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2xy}>=2\) (3)
Từ (1) , (2) và (3) , suy ra : \(K>=4+2=6\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x^2+y^2=2xy\\x=y\\x+y=1\end{cases}}\)
\(\Rightarrow x=y=\frac{1}{2}\)
Vậy Min\(K=6\)khi \(x=y=\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 1 2022
\(x+y=xy\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}=1\)
Đặt \(\left(\dfrac{1}{x};\dfrac{1}{y}\right)=\left(a;b\right)\Rightarrow a+b=1\) \(\Rightarrow a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2=\dfrac{1}{2}\)
\(P=\dfrac{a^2}{1+a-a^2}+\dfrac{b^2}{1+b-b^2}\ge\dfrac{\left(a+b\right)^2}{2+a+b-\left(a^2+b^2\right)}=\dfrac{1}{3-\left(a^2+b^2\right)}\ge\dfrac{1}{3-\dfrac{1}{2}}=\dfrac{2}{5}\)
Dấu "=" xảy ra khi \(x=y=2\)
24 tháng 11 2021
\(1,\) Áp dụng BĐT: \(x^2+y^2\ge\dfrac{\left(x+y\right)^2}{2}\text{ và }\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Dấu \("="\Leftrightarrow x=y\)
\(A=\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\\ A\ge\dfrac{1}{2}\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2+17\ge\dfrac{1}{2}\left(1+\dfrac{4}{a+b}\right)^2+17=\dfrac{25}{2}+17=\dfrac{59}{2}\\ \text{Dấu }"="\Leftrightarrow\left\{{}\begin{matrix}a+\dfrac{1}{a}=b+\dfrac{1}{b}\\a+b=1\end{matrix}\right.\Leftrightarrow a=b=\dfrac{1}{2}\)
24 tháng 11 2021
\(2,\text{Đặt }A=\dfrac{xy}{z}+\dfrac{yz}{x}+\dfrac{xz}{y}\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(\dfrac{xy^2z}{xz}+\dfrac{xyz^2}{xy}+\dfrac{x^2yz}{yz}\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+2\left(x^2+y^2+z^2\right)\\ \Leftrightarrow A^2=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}+6\)
Áp dụng Cosi: \(\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}\ge2y^2\)
CMTT: \(\left\{{}\begin{matrix}\dfrac{y^2z^2}{x^2}+\dfrac{x^2z^2}{y^2}\ge2z^2\\\dfrac{x^2y^2}{z^2}+\dfrac{x^2z^2}{y^2}\ge2x^2\end{matrix}\right.\)
Cộng VTV \(\Leftrightarrow A^2\ge2\left(x^2+y^2+z^2\right)+6=12\\ \Leftrightarrow A\ge2\sqrt{3}\)
Dấu \("="\Leftrightarrow x=y=z=1\)
23 tháng 1 2021
1) Áp dụng bất đẳng thức AM - GM và bất đẳng thức Schwarz:
\(P=\dfrac{1}{a}+\dfrac{1}{\sqrt{ab}}\ge\dfrac{1}{a}+\dfrac{1}{\dfrac{a+b}{2}}\ge\dfrac{4}{a+\dfrac{a+b}{2}}=\dfrac{8}{3a+b}\ge8\).
Đẳng thức xảy ra khi a = b = \(\dfrac{1}{4}\).
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 1 2021
2.
\(4=a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Rightarrow a+b\le2\sqrt{2}\)
Đồng thời \(\left(a+b\right)^2\ge a^2+b^2\Rightarrow a+b\ge2\)
\(M\le\dfrac{\left(a+b\right)^2}{4\left(a+b+2\right)}=\dfrac{x^2}{4\left(x+2\right)}\) (với \(x=a+b\Rightarrow2\le x\le2\sqrt{2}\) )
\(M\le\dfrac{x^2}{4\left(x+2\right)}-\sqrt{2}+1+\sqrt{2}-1\)
\(M\le\dfrac{\left(2\sqrt{2}-x\right)\left(x+4-2\sqrt{2}\right)}{4\left(x+2\right)}+\sqrt{2}-1\le\sqrt{2}-1\)
Dấu "=" xảy ra khi \(x=2\sqrt{2}\) hay \(a=b=\sqrt{2}\)
3. Chia 2 vế giả thiết cho \(x^2y^2\)
\(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\ge\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\)
\(\Rightarrow0\le\dfrac{1}{x}+\dfrac{1}{y}\le4\)
\(A=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}-\dfrac{1}{xy}\right)=\left(\dfrac{1}{x}+\dfrac{1}{y}\right)^2\le16\)
Dấu "=" xảy ra khi \(x=y=\dfrac{1}{2}\)
13 tháng 12 2021
\(P=\sum\dfrac{1}{x+y+1}\ge\dfrac{9}{2\left(x+y+z\right)+3}=\dfrac{9}{2.1+3}=\dfrac{9}{5}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Xét \(B=\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\)
Áp dụng bất đẳng thức: \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{\left(a+b\right)^2}\), ta có:
\(\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}\ge\dfrac{4}{x^2+2xy+y^2}=\dfrac{4}{\left(x+y\right)^2}=\dfrac{4}{1^2}=4\)
\(\Rightarrow B\ge4\)
Ta có:
\(\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow1\ge4xy\)
\(\Leftrightarrow\dfrac{1}{2xy}\ge\dfrac{4xy}{2xy}=2\) (x,y>0)
Khi đó:
\(A=B+\dfrac{1}{2xy}\ge4+2=6\)
Dấu "=" xảy ra \(\Leftrightarrow\) \(x=y=\dfrac{1}{2}\)
\(A=\dfrac{1}{x^2+y^2}+\dfrac{1}{xy}\\ =\dfrac{1}{x^2+y^2}+\dfrac{1}{2xy}+\dfrac{2}{4xy}\\ \overset{AM-GM}{\ge}\dfrac{4}{x^2+y^2+2xy}+\dfrac{2}{\left(x+y\right)^2}\\ =\dfrac{4}{\left(x+y\right)^2}+\dfrac{2}{\left(x+y\right)^2}=4+2=6\)
Dấu "=" xảy ra khi \(:\left\{{}\begin{matrix}x^2+y^2=2xy\\x=y\end{matrix}\right.\Leftrightarrow x=y\)
Vậy \(A_{Min}=6\) khi \(x=y\)