NP
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
4 tháng 10 2019
\(x^2+y^2+z^2+2xyz=1\)
\(\Leftrightarrow2xyz=1-x^2-y^2-z^2\)
\(\Rightarrow P=xy+yz+xz-2xyz=xy+yz+xz+x^2+y^2+z^2-1\)
\(\Rightarrow2P=\left(x+y\right)^2+\left(y+z\right)^2+\left(z+x\right)^2-2\ge1\)
\(\Rightarrow P\ge\frac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)
AJ
1
NV
Nguyễn Việt Lâm
Giáo viên
16 tháng 2 2020
Bài này x;y;z phải dương chứ nhỉ? Có dấu "=" ở số 0 thế kia thì bối rối quá
Dự đoán dấu "=" xảy ra tại \(x=y=z=\frac{1}{2}\)
Theo nguyên lý Dirichlet, trong 3 số x;y;z luôn tồn tại 2 số nằm cùng phía so với \(\frac{1}{2}\) ; giả sử đó là x và y
\(\Rightarrow\left(x-\frac{1}{2}\right)\left(y-\frac{1}{2}\right)\ge0\Leftrightarrow\frac{1}{2}\left(x+y\right)-xy\le\frac{1}{4}\)
\(\Leftrightarrow x+y-2xy\le\frac{1}{2}\)
Mặt khác:
\(1=2xyz+x^2+y^2+z^2\ge2xyz+2xy+z^2=2xy\left(1+z\right)+z^2\)
\(\Rightarrow1-z^2\ge2xy\left(1+z\right)\Leftrightarrow\left(1-z\right)\left(1+z\right)\ge2xy\left(1+z\right)\)
\(\Leftrightarrow1-z\ge2xy\Rightarrow xy\le\frac{1-z}{2}\)
\(\Rightarrow P=xy+z\left(x+y-2xy\right)\le\frac{1-z}{2}+\frac{z}{2}=\frac{1}{2}\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{2}\)
20 tháng 5 2016
\(P+3=\frac{xy}{1+x+y}+1+\frac{yz}{1+y+z}+1+\frac{xz}{1+x+z}+1\)
\(\frac{xy}{1+x+y}+1=\frac{\left(x+1\right)\left(y+1\right)}{1+x+y}\)
\(P+3=\left(x+1\right)\left(y+1\right)\left(z+1\right)\left(\frac{1}{\left(z+1\right)\left(x+y+1\right)}+\frac{1}{\left(y+1\right)\left(x+z+1\right)}+\frac{1}{\left(x+1\right)\left(y+z+1\right)}\right)\)
\(P+3\ge\left(xyz+xy+xz+yz+1\right)\left(\frac{9}{xy+xz+x+y+z+1+xy+yz+x+y+z+1+xz+yz+x+y+z+1}\right)\)
20 tháng 5 2016
dòng cuối cùng sai, sửa :
\(P+3\ge\left(xyz+xy+xz+yz+1\right)\left(\frac{9}{xy+xz+x+y+z+1+xy+yz+x+y+z+1+xz+yz+x+y+z+1}\right)\)
\(P+3\ge\left(3xyz+xy+xz+yz\right)\left(\frac{9}{2\left(3xyz+xy+xz+yz\right)}\right)=\frac{9}{2}\)
\(P\ge\frac{3}{2}\)
dấu "=" xảy ra <=> x=y=z=\(\frac{1+\sqrt{3}}{2}\)
D
1
HF
18 tháng 8 2020
+) \(P=\sqrt{1-x^2}+\sqrt{1-y^2}+\sqrt{1-z^2}\)
\(\le\frac{1-x^2+\frac{3}{4}}{\sqrt{3}}+\frac{1-y^2+\frac{3}{4}}{\sqrt{3}}+\frac{1-z^2+\frac{3}{4}}{\sqrt{3}}\)
\(=\frac{\frac{21}{4}-x^2-y^2-z^2}{\sqrt{3}}\)
+) \(1=xy+yz+xz+2xyz\le\frac{\left(x+y+z\right)^2}{3}+\frac{2\left(x+y+z\right)^3}{27}\)
Đặt \(a=x+y+z\), ta được \(2a^3+9a^2-27\ge0\Leftrightarrow\left(2a-3\right)\left(a+3\right)^2\ge0\Rightarrow a\ge\frac{3}{2}\)
+) \(A=x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}=\frac{\frac{9}{4}}{3}=\frac{3}{4}\)
+) \(P\ge\frac{\frac{21}{4}-A}{\sqrt{3}}=\frac{\frac{21}{4}-\frac{3}{4}}{\sqrt{3}}=\frac{9}{2\sqrt{3}}=\frac{3\sqrt{3}}{2}\)
Dấu = xảy ra khi x = y = z = 1/2
KK
0
9 tháng 4 2017
\(P=\dfrac{xy}{1+x+y}+\dfrac{yz}{1+y+z}+\dfrac{xz}{1+z+x}\)
\(P+3=\dfrac{xy}{1+x+y}+1+\dfrac{yz}{1+y+z}+1+\dfrac{xz}{1+z+x}+1\)
\(P+3=\dfrac{\left(x+1\right)\left(y+1\right)}{1+x+y}+\dfrac{\left(y+1\right)\left(z+1\right)}{1+y+z}+\dfrac{\left(x+1\right)\left(z+1\right)}{1+z+x}\)
\(P+3=\dfrac{\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(1+x+y\right)\left(z+1\right)}+\dfrac{\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(x+1\right)\left(1+y+z\right)}+\dfrac{\left(x+1\right)\left(y+1\right)\left(z+1\right)}{\left(y+1\right)\left(1+z+x\right)}\)
\(P+3=\left(x+1\right)\left(y+1\right)\left(z+1\right)\left[\dfrac{1}{\left(1+x+y\right)\left(z+1\right)}+\dfrac{1}{\left(x+1\right)\left(1+y+z\right)}+\dfrac{1}{\left(y+1\right)\left(1+z+x\right)}\right]\)
\(\ge\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\left(1+x+y\right)\left(z+1\right)+\left(x+1\right)\left(1+y+z\right)+\left(y+1\right)\left(1+z+x\right)}\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\text{ }2xy+2yz+2xz+3x+3y+3z+3}\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\text{ }2xy+2yz+2xz+3\cdot2xyz}\)
\(=\left(x+1\right)\left(y+1\right)\left(z+1\right)\cdot\dfrac{9}{\text{ }2\left(xy+yz+xz+3xyz\right)}\)
Lại có:
\(\left(x+1\right)\left(y+1\right)\left(z+1\right)=xyz+xy+yz+xz+x+y+z+1\)
\(=xyz+xy+yz+xz+2xyz=xy+yz+xz+3xyz\)
\(\Rightarrow P+3\ge\left(xy+yz+xz+3xyz\right)\cdot\dfrac{9}{2\left(xy+yz+xz+3xyz\right)}\)
\(\Rightarrow P+3\ge\dfrac{9}{2}\Rightarrow P\ge\dfrac{9}{2}-3=\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(x=y=z=\dfrac{1+\sqrt{3}}{2}\)
VT
0
Ta có:
P=\(\left(X^2+y^2+z^2+2xyz\right)-\left(X^2+y^2+z^2+4xyz-xy-yz-xz\right)\) xz)
= 1-\(\left(x^2+y^2+z^2+4xyz-xy-yz-xz\right)\)
=> P \(\le\)1
Vậy MaxP=1