tìm GTNN
P=\(\frac{1}{1+xy}+\frac{1}{1+xz}+\frac{1}{1+yz}\)
biết x,y,z >0 và \(x^2+y^2+z^2\)<=3
K
Khách
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SX
1
TN
2 tháng 6 2017
Câu hỏi của Minh Hà Tuấn - Toán lớp 9 - Học toán với OnlineMath
TV
Cho x,y,z>0 thỏa \(x^2+y^2+z^2< =3\)
Tìm GTNN của P= \(\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+xz}\)
2
PA
2 tháng 8 2017
Áp dụng BĐT Cauchy Shwarz, ta có:
\(M=\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+xz}\)
\(\ge\frac{\left(1+1+1\right)^2}{1+1+1+xy+yz+xz}\)
\(\ge\frac{9}{3+x^2+y^2+z^2}\)
\(=\frac{3}{2}\)
Dấu "=" xảy ra khi x = y = z = 1
2 tháng 8 2017
\(P=\frac{1}{1+xy}+\frac{1}{1+yz}+\frac{1}{1+zx}\ge\frac{9}{3+xy+yz+zx}\)
\(\ge\frac{9}{3+x^2+y^2+z^2}\ge\frac{9}{3+3}=\frac{3}{2}\)
Dấu = xảy ra khi \(x=y=z=1\)
KM
1
I
13 tháng 4 2020
Áp dung BĐT \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\left(a,b,c>0\right)\)
\(=>x,y,z>0\left(taco\right)\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge\frac{9}{xy+yz+xz}\)
\(=>P\ge\frac{1}{x^2+y^2+z^2}+\frac{9}{xy+yz+xz}\)
\(=>P\ge\left(\frac{1}{x^2+y^2+z^2}+\frac{1}{xy+yz+zx}+\frac{1}{xy+yz+zx}\right)+\frac{7}{xy+yz+xz}\)
\(\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}+\frac{7}{xy+yz+zx}\)
\(=\frac{9}{\left(x+y+z\right)^2}+\frac{7}{xy+yz+xz}\ge\frac{9}{\left(x+y+z\right)^2}+\frac{21}{\left(x+y+z\right)^2}\ge30\)
do \(3\left(xy+yz+zx\right)\le\left(x+y+z\right)^2and\left(x+y+z=1\right)\)
dấu = xảy ra khi x=y=z=1/3
zậy...........
LT
1
AH
Akai Haruma
Giáo viên
11 tháng 3 2019
Lời giải:
Vì \(x,y,z\in [0;1]\Rightarrow xy; yz,xz\geq xyz\)
\(\Rightarrow P=\frac{x}{1+yz}+\frac{y}{1+xz}+\frac{z}{xy+1}\leq \frac{x}{1+xyz}+\frac{y}{1+xyz}+\frac{z}{1+xyz}=\frac{x+y+z}{xyz+1}(*)\)
\(x,y,z\in [0;1]\Rightarrow \left\{\begin{matrix} (x-1)(y-1)\geq 0\\ (xy-1)(z-1)\geq 0\end{matrix}\right.\)
\(\Rightarrow \left\{\begin{matrix} xy+1\geq x+y\\ xyz+1\geq xy+z\end{matrix}\right.\)
\(\Rightarrow xyz+2+xy\geq x+y+z+xy\)
\(\Leftrightarrow xyz+2\geq x+y+z\)
Mà: \(xyz+2\leq 2xyz+2=2(xyz+1)\)
\(\Rightarrow x+y+z\leq 2(xyz+1)(**)\)
Từ \((*); (**)\Rightarrow P\leq \frac{2(xyz+1)}{xyz+1}=2\) (đpcm)
Dấu "=" xảy ra khi \((x,y,z)=(1,1,0)\)
KK
21 tháng 1 2017
Áp dụng BĐT Cô - si cho 3 bộ số không âm
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{xyz\left(xy+1\right)^2\left(yz+1\right)^2\left(xz+1\right)^2}{x^2y^2z^2\left(yz+1\right)\left(xz+1\right)\left(xy+1\right)}}=3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
Xét \(3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(=3\sqrt[3]{\left(\frac{xy+1}{x}\right)\left(\frac{yz+1}{y}\right)\left(\frac{xz+1}{z}\right)}\)
\(=3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\)
Áp dụng BĐT Cô - si
\(\Rightarrow\left\{\begin{matrix}y+\frac{1}{x}\ge2\sqrt{\frac{y}{x}}\\z+\frac{1}{y}\ge2\sqrt{\frac{z}{y}}\\x+\frac{1}{z}\ge2\sqrt{\frac{x}{z}}\end{matrix}\right.\)
\(\Rightarrow\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)\ge8\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge3\sqrt[3]{8}\)
\(\Rightarrow3\sqrt[3]{\left(y+\frac{1}{x}\right)\left(z+\frac{1}{y}\right)\left(x+\frac{1}{z}\right)}\ge6\)
\(\Leftrightarrow3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\ge6\)
Mà \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge3\sqrt[3]{\frac{\left(xy+1\right)\left(yz+1\right)\left(xz+1\right)}{xyz}}\)
\(\Rightarrow\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}\ge6\)
Vậy GTNN của \(\frac{z\left(xy+1\right)^2}{y^2\left(yz+1\right)}+\frac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\frac{y\left(xz+1\right)^2}{x^2\left(xy+1\right)}=6\)
\(P=\frac{1}{1+xy}+\frac{1}{1+xz}+\frac{1}{1+yz}\ge\frac{9}{3+xy+xz+yz}\)
Lại có :\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow xy+yz+zx\le x^2+y^2+z^2\le3\)
\(\Rightarrow P\ge\frac{9}{3+3}=1.5\)
Dấu bằng xảy ra khi x=y=z=1