\(\frac{1}{\text{x}^2+yz}+\frac{1}{y^2+\text{x}z}+\frac{1}{z^2+\text{x}y}\le\frac{1}{2}\left(\frac{1}{\text{x}y}+\frac{1}{yz}+\frac{1}{\text{x}z}\right)\)
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ML
8 tháng 8 2016
+\(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)
+\(3+2\left(xy+yz+zx\right)=x^2+y^2+z^2+2\left(xy+yz+zx\right)=\left(x+y+z\right)^2\le9\)
\(\Rightarrow B=\frac{1}{1+\sqrt{3+2\left(xy+yz+zx\right)}}\ge\frac{1}{1+3}=\frac{1}{4}\)
+\(A=\frac{x^2}{y+2z}+\frac{y^2}{z+2x}+\frac{z^2}{x+2y}=\frac{x^4}{x^2y+2zx^2}+\frac{y^4}{y^2z+2xy^2}+\frac{z^4}{z^2x+2yz^2}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{x^2y+y^2z+z^2x+2\left(xy^2+yz^2+zx^2\right)}\)
Áp dụng bđt Bunhiacopxki
\(x^2y+y^2z+z^2x=x.xy+y.yz+z.zx\le\sqrt{x^2+y^2+z^2}.\sqrt{x^2y^2+y^2z^2+z^2x^2}\)
\(\le\sqrt{x^2+y^2+z^2}.\sqrt{\frac{\left(x^2+y^2+z^2\right)^2}{3}}=3\)
(áp dụng \(ab+bc+ca\le\frac{\left(a+b+c\right)^2}{3}\))
Tương tự: \(xy^2+yz^2+zx^2\le3\)
\(\Rightarrow B\ge\frac{3^2}{3+2.3}=1\)
\(VT=A+B\ge1+\frac{1}{4}=\frac{5}{4}=VP\)
TD
0
NV
Nguyễn Việt Lâm
Giáo viên
9 tháng 8 2020
\(P=\frac{x\left(x+y+z\right)+yz}{y+z}+\frac{y\left(x+y+z\right)+zx}{z+x}+\frac{z\left(x+y+z\right)+xy}{x+y}\)
\(P=\frac{\left(x+y\right)\left(x+z\right)}{y+z}+\frac{\left(x+y\right)\left(y+z\right)}{z+x}+\frac{\left(x+z\right)\left(y+z\right)}{x+y}\)
\(P\ge\left(x+y\right)+\left(y+z\right)+\left(z+x\right)=2\left(x+y+z\right)=2\)
Dấu "=" xảy ra khi \(x=y=z=\frac{1}{3}\)
AH
Akai Haruma
Giáo viên
19 tháng 1 2017
Bài này đơn giản thôi :))
\(\text{HPT}\Leftrightarrow \left\{\begin{matrix} \frac{x+y}{xy}=\frac{3}{2}\\ \frac{y+z}{yz}=\frac{2}{3}\\ \frac{x+z}{xz}=\frac{7}{6}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}=\frac{3}{2}\\ \frac{1}{y}+\frac{1}{z}=\frac{2}{3}\\ \frac{1}{x}+\frac{1}{z}=\frac{7}{6}\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} \frac{2}{x}=\frac{3}{2}+\frac{7}{6}-\frac{2}{3}\\ \frac{2}{y}=\frac{3}{2}+\frac{2}{3}-\frac{7}{6}\\ \frac{2}{z}=\frac{2}{3}+\frac{7}{6}-\frac{3}{2}\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=2\\ z=6\end{matrix}\right.\)
Vậy $(x,y,z)=(1,2,6)$ là nghiệm của hệ phương trình
19 tháng 8 2020
Đặt \(P=\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Do x,y,z là các số thực dương nên ta biến đổi \(P=\frac{1}{\sqrt{1+\frac{1}{x^2}}}+\frac{1}{\sqrt{1+\frac{1}{y^2}}}+\frac{1}{\sqrt{1+\frac{1}{z^2}}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Đặt \(a=\frac{1}{x^2};b=\frac{1}{y^2};c=\frac{1}{z^2}\left(a,b,c>0\right)\)thì \(xy+yz+zx=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}=1\)và \(P=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}+a+b+c\)
Biến đổi biểu thức P=\(\left(\frac{1}{2\sqrt{a+1}}+\frac{1}{2\sqrt{a+1}}+\frac{a+1}{16}\right)+\left(\frac{1}{2\sqrt{b+1}}+\frac{1}{2\sqrt{b+1}}+\frac{b+1}{16}\right)\)\(+\left(\frac{1}{2\sqrt{c+1}}+\frac{1}{2\sqrt{c+1}}+\frac{c+1}{16}\right)+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{b}-\frac{3}{16}\)
Áp dụng Bất Đẳng Thức Cauchy ta có
\(P\ge3\sqrt[3]{\frac{a+1}{64\left(a+1\right)}}+3\sqrt[3]{\frac{b+1}{64\left(b+1\right)}}+3\sqrt[3]{\frac{c+1}{64\left(c+1\right)}}+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{16}-\frac{3}{16}\)
\(=\frac{33}{16}+\frac{15}{16}\left(a+b+c\right)\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{abc}\)
Mặt khác ta có \(1=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge3\sqrt[3]{\frac{1}{abc}}\Leftrightarrow abc\ge27\)
\(\Rightarrow P\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{27}=\frac{33}{16}+\frac{15}{16}\cdot9=\frac{21}{2}\)
Dấu "=" xảy ra khi a=b=c hay \(x=y=z=\frac{\sqrt{3}}{3}\)
đk: x;y;z dương nhé
áp dụng bđt cosi ta có:
\(x^2+yz>=2\sqrt{x^2yz}=2x\sqrt{yz};y^2+xz>=2\sqrt{y^2xz}=2y\sqrt{xz};z^2+xy=2\sqrt{z^2xy}=2z\sqrt{xy}\)
\(\Rightarrow\frac{1}{x^2+yz}< =\frac{1}{2x\sqrt{yz}};\frac{1}{y^2+xz}< =\frac{1}{2y\sqrt{xz}};\frac{1}{z^2+xy}< =\frac{1}{2z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}< =\frac{1}{2x\sqrt{yz}}+\frac{1}{2y\sqrt{xz}}+\frac{1}{2z\sqrt{xy}}=\frac{1}{2}\left(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\left(1\right)\)
áp dụng bđt cosi ta có:
\(\frac{1}{xy}+\frac{1}{xz}>=2\cdot\sqrt{\frac{1}{xy}\cdot\frac{1}{xz}}=\frac{2}{x\sqrt{yz}};\frac{1}{xy}+\frac{1}{yz}>=2\cdot\sqrt{\frac{1}{xy}\cdot\frac{1}{yz}}=\frac{2}{y\sqrt{xz}};\)
\(\frac{1}{yz}+\frac{1}{xz}>=2\cdot\sqrt{\frac{1}{yz}\cdot\frac{1}{xz}}=\frac{2}{z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{xy}+\frac{1}{xz}+\frac{1}{xy}+\frac{1}{yz}+\frac{1}{yz}+\frac{1}{xz}=\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}>=\frac{2}{x\sqrt{yz}}+\frac{2}{y\sqrt{xz}}+\frac{2}{z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}>=\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\)
\(\Rightarrow\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)>=\frac{1}{2}\left(\frac{1}{x\sqrt{yz}}+\frac{1}{y\sqrt{xz}}+\frac{1}{z\sqrt{xy}}\right)\left(2\right)\)
từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}>=\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\left(đpcm\right)\)
dấu = xảy ra khi x=y=z
nhầm từ \(\left(1\right);\left(2\right)\Rightarrow\frac{1}{x^2+yz}+\frac{1}{y^2+xz}+\frac{1}{z^2+xy}< =\frac{1}{2}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)