Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a,\(A\ge\frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\ge\frac{9}{\sqrt{3\left(x+y+z\right)}}=3\)=3
MInA=3<=>x=y=z=1
b)dùng cô si đi(đề thi chuyên bình phước năm 2016-2017)
\(x+y\le\sqrt{2\left(x^2+y^2\right)}=\sqrt{2}\Rightarrow\frac{1}{x+y}\ge\frac{\sqrt{2}}{2}\)
\(P=x+y+\frac{x}{y}+\frac{y}{x}+\frac{1}{x}+\frac{1}{y}+2\ge x+y+2\sqrt{\frac{xy}{xy}}+\frac{4}{x+y}+2\)
\(P\ge x+y+\frac{2}{x+y}+\frac{2}{x+y}+4\ge2\sqrt{\frac{2\left(x+y\right)}{x+y}}+2.\frac{\sqrt{2}}{2}+4=4+3\sqrt{2}\)
\(\Rightarrow P_{min}=4+3\sqrt{2}\) khi \(x=y=\frac{1}{\sqrt{2}}\)
Bài làm:
Ta có: \(M=\left(x^2+\frac{1}{y^2}\right)\left(y^2+\frac{1}{x^2}\right)\)
\(=x^2y^2+2+\frac{1}{x^2y^2}\)
\(=\left(x^2y^2+\frac{1}{256x^2y^2}\right)+\frac{255}{256x^2y^2}+2\)
Mà \(xy\le\frac{\left(x+y\right)^2}{4}=\frac{1}{4}\Rightarrow x^2y^2\le\frac{1}{16}\)
Thay vào ta tính được:
\(M\ge2\sqrt{x^2y^2\cdot\frac{1}{256x^2y^2}}+\frac{255}{256\cdot\frac{1}{16}}+2\)
\(=\frac{1}{8}+\frac{255}{16}+2=\frac{289}{16}\)
Dấu "=" xảy ra khi: \(x=y=\frac{1}{2}\)
Vậy \(Min_M=\frac{289}{16}\Leftrightarrow x=y=\frac{1}{2}\)
Đánh máy xong hết lại bấm hủy-.-
ta có
\(0\le\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\left(\forall x,y,z>0\right)\)
\(\Leftrightarrow2xy+2yz+2zx\le2\left(x^2+y^2+z^2\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)\)(1)
dấu = xảy ra khi
\(x=y=z=0\)
theo giả thiết ta có
\(x\left(x+1\right)+y\left(y+1\right)+z\left(z+1\right)\le18\)
\(\Leftrightarrow x^2+y^2+z^2\le18-\left(x+y+z\right)\left(2\right)\)
từ (1) zà (2) suy ra
\(\left(x+y+z\right)^2\le54-3\left(x+y+z\right)\)
\(\Leftrightarrow\left(x+y+z\right)^2+3\left(x+y+z\right)-54\le0\)
\(\Leftrightarrow\left(x+y+z-6\right)\left(x+y+z+9\right)\le0\)
\(\Leftrightarrow0< x+y+z\le6\left(do\left(x+y+z>0;9>0\right)\right)\)
áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\)ta có
\(P=\frac{1}{x+y+1}+\frac{1}{y+z+1}+\frac{1}{z+x+1}\ge\frac{9}{2\left(x+y+z\right)+3}\ge\frac{9}{2.6+3}=\frac{3}{5}\)
Dấu = xảy ra khi zà chỉ khi
\(\hept{\begin{cases}x+y+1=y+z+1=z+x+1\\x+y+z=6\end{cases}=>x=y=z=2}\)
zậy MinP= 3/5 khi x=y=z=2
Ta có : x(x + 1) + y (y+1 ) + z(z + 1) \(\le18\)
<=> x2 + y2 + z2 + ( x + y + z ) \(\le18\)
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\Rightarrow3\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2\)
=> 54 \(\ge\)( x + y+z)2 + 3(x + y + z)
<=> -9 \(\le\)x + y + z \(\le\)6
=> 0 \(\le\)x+y+z \(\le\)6
\(\frac{1}{x+y+1}+\frac{x+y+1}{25}\ge\frac{2}{5}\)
\(\frac{1}{y+z+1}+\frac{y+z+1}{25}\ge\frac{2}{5}\)
\(\frac{1}{z+x+1}+\frac{z+x+1}{25}\ge\frac{2}{5}\)
=> \(P+\frac{2\left(x+y+z\right)+3}{25}\ge\frac{6}{5}\)
=> P \(\ge\frac{27}{25}-\frac{2}{25}\left(x+y+z\right)\ge\frac{15}{25}=\frac{3}{5}\)
Dấu " =" xảy ra khi :
\(\hept{\begin{cases}x=y=z>0;x+y+z=6\\\left(x+y+1\right)^2=\left(y+z+1\right)^2=\left(z+x+1\right)^2=25\end{cases}\Leftrightarrow x=y=z=2}\)
Vậy GTNN của P là \(\frac{3}{5}\)khi x = y =z =2