cho x,y,z dương thỏa \(x^2+y^2+z^2=3\)
C/M \(\dfrac{x}{3-yz}+\dfrac{y}{3-zx}+\dfrac{z}{3-xy}\le\dfrac{3}{2}\)
Cho ba số thực dương x, y, z thỏa mãn: xy+yz+zx=2017. chứng minh : \(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{3}{2}\)
Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)
Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)
\(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
Cộng vế với vế ta có:
\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)
\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)
cho x,y,z là các số thực dương thỏa mãn : xy+yz+zx=2016
c/m : \(\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{y^2+2016}}+\sqrt{\dfrac{xz}{z^2+2016}}\le\dfrac{3}{2}\)
\(VT=\sqrt{\dfrac{yz}{x^2+xy+yz+xz}}+\sqrt{\dfrac{xy}{y^2+xy+yz+xz}}+\sqrt{\dfrac{xz}{z^2+xy+yz+xz}}\)
\(VT=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{xy}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xz}{\left(x+z\right)\left(y+z\right)}}\)
Áp dụng bất đẳng thức Cauchy - Schwarz
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\\\sqrt{\dfrac{xy}{\left(y+z\right)\left(x+y\right)}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{y}{y+z}}{2}\\\sqrt{\dfrac{xz}{\left(x+z\right)\left(y+z\right)}}\le\dfrac{\dfrac{x}{x+z}+\dfrac{z}{y+z}}{2}\end{matrix}\right.\)
\(\Rightarrow VT\le\dfrac{\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}\right)+\left(\dfrac{y}{y+z}+\dfrac{z}{y+z}\right)+\left(\dfrac{z}{x+z}+\dfrac{x}{x+z}\right)}{2}\)
\(\Rightarrow VT\le\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{x+z}{x+z}}{2}=\dfrac{3}{2}\)
\(\Leftrightarrow\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{y^2+2016}}+\sqrt{\dfrac{xz}{z^2+2016}}\le\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(x=y=z=4\sqrt{42}\)
Sửa đề:\(\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{z^2+2016}}+\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{3}{2}\)
Giải
Ta có:
\(\sqrt{\dfrac{xy}{z^2+2016}}=\sqrt{\dfrac{xy}{z^2+xy+xz+yz}}=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{\dfrac{xy}{z^2+2016}}=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)\)
Tương tự cho 2 BĐT còn lại ta có:
\(\sqrt{\dfrac{yz}{x^2+2016}}\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right);\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{z}{y+z}\right)\)
Cộng theo vế 3 BĐT trên ta có:
\(\Sigma\sqrt{\dfrac{xy}{z^2+2016}}\le\dfrac{1}{2}\Sigma\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{1}{2}\Sigma\left(\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)=\dfrac{3}{2}\)
Đẳng thức xảy ra khi \(x=y=z=4\sqrt{42}\)
xí bài này nhé, lát nữa hoặc mai giải
Cho x, y, z >0 thỏa mãn x + y + z = 1
CMR: \(\sqrt{\dfrac{xy}{xy+z}}+\sqrt{\dfrac{yz}{yz+x}}+\sqrt{\dfrac{zx}{zx+y}}\le\dfrac{3}{2}\)
Lời giải:
Áp dụng BĐT AM-GM:
\(\sqrt{\frac{xy}{xy+z}}=\sqrt{\frac{xy}{xy+z(x+y+z)}}=\sqrt{\frac{xy}{(z+x)(z+y)}}\leq \frac{1}{2}\left(\frac{x}{x+z}+\frac{y}{z+y}\right)\)
Hoàn toàn tương tự với các phân thức còn lại suy ra:
\(\sum \sqrt{\frac{xy}{xy+z}}\leq \frac{1}{2}\left(\frac{x+z}{x+z}+\frac{y+z}{y+z}+\frac{x+y}{x+y}\right)=\frac{3}{2}\)
Ta có đpcm.
Dấu "=" xảy ra khi $x=y=z=\frac{1}{3}$
Cho các số thực dương x, y, z thỏa mãn x2 + y2 + z2 = 3. CMR \(\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}\ge xy+yz+zx\)
Bài này cũng dễ mà:
Áp dụng BĐT Cô-si, ta có:
\(y+z+1\ge3\sqrt[3]{yz}\)
\(\Rightarrow\)\(\dfrac{y+z+1}{3}\ge\sqrt[3]{yz}\)
\(\Rightarrow\)\(\dfrac{x}{\sqrt[3]{yz}}\ge\dfrac{3x}{y+z+1}\)
\(\Rightarrow\)\(\sum\dfrac{x}{\sqrt[3]{yz}}\ge\sum\dfrac{3x}{y+z+1}\)
Mà \(\sum\dfrac{3x}{y+z+1}=\sum\dfrac{3x^2}{xy+xz+x}\)
Áp dụng BĐT Cauchy -Schwaz:
\(\sum\dfrac{3x^2}{xy+xz+x}\ge\dfrac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
Mà:
\(xy+yz+xz\le x^2+y^2+z^2\)(BĐT phụ)
\(\Rightarrow\)\(2\left(xy+yz+xz\right)\le2\left(x^2+y^2+z^2\right)=6\)
Áp dụng BĐT Bunhicopski:
\(\left(x+y+z\right)^2\le3\left(x^2+y^2+z^2\right)=9\)
\(\Rightarrow x+y+z\le3\)
\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le6+3=9\)
\(\Rightarrow\)\(\dfrac{3\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{3\left(x+y+z\right)^2}{9}\ge\dfrac{\left(x+y+z\right)^2}{3}\ge xy+yz+xz\left(ĐPCM\right)\)
Dấu "=" xảy ra \(\Leftrightarrow\)x=y=z=1
@Lightning Farron vào thể hiện đẳng cấp đi anh zai :))
Cho x,y,z dương thỏa mãn x+y+z=1. Tìm minP = \(\dfrac{3}{xy+yz+zx}+\dfrac{2}{x^2+y^2+z^2}\)
\(P=\dfrac{6}{2xy+2yz+2zx}+\dfrac{2}{x^2+y^2+z^2}\ge\dfrac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=8+4\sqrt{3}\)
Cho các số dương x;y;z. CMR:
\(\dfrac{xy}{x^2+yz+zx}+\dfrac{yz}{y^2+zx+xy}+\dfrac{zx}{z^2+xy+yz}\le\dfrac{x^2+y^2+z^2}{xy+yz+zx}\)
Cho x;y;z>0 thỏa mãn \(x^2+y^2+z^2=3\)
chứng minh: \(\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{zx}}+\dfrac{z}{\sqrt[3]{xy}}\ge xy+yz+zx\)
Ta có : Áp dụng BĐT Cauchy ba số ở mẫu ta được
\(\dfrac{x}{\sqrt[3]{yz}}+\dfrac{y}{\sqrt[3]{xz}}+\dfrac{z}{\sqrt[3]{xy}}\ge\dfrac{x}{\dfrac{y+z+1}{3}}+\dfrac{y}{\dfrac{x+z+1}{3}}+\dfrac{z}{\dfrac{x+y+1}{3}}=\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\)Thấy: \(xy+yz+xz\le\dfrac{\left(x+y+z\right)^2}{3}\left(?!\right)\)
Ta phải chứng minh:
\(\dfrac{3x}{y+z+1}+\dfrac{3y}{x+z+1}+\dfrac{3z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{3}\)
\(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}\ge\dfrac{\left(x+y+z\right)^2}{9}\)
Mà \(\dfrac{x}{y+z+1}+\dfrac{y}{x+z+1}+\dfrac{z}{x+y+1}=\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\)
Theo C.B.S
\(\dfrac{x^2}{xy+xz+x}+\dfrac{y^2}{xy+yz+y}+\dfrac{z^2}{xz+yz+z}\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\)
Phải chứng minh
\(\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{\left(x+y+z\right)^2}{9}\)
\(\Leftrightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)
Ta có : \(xy+yz+xz\le x^2+y^2+z^2=3\)
Theo C.B.S : \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}=3\)
\(\Rightarrow2\left(xy+yz+xz\right)+x+y+z\le9\)
\(\Rightarrow\dfrac{1}{2\left(xy+yz+xz\right)+x+y+z}\ge\dfrac{1}{9}\)
=> ĐPCM
cho 3 số thực dương x,y,z thỏa mãn x+y+z=3.chứng minh:
\(\dfrac{x}{y^3+xy}+\dfrac{y}{z^3+zx}+\dfrac{z}{x^3+xy}\)≥\(\dfrac{3}{2}\)
1, Cho các số x,y,z không âm. \(\ne\)0. thỏa mãn: \(\dfrac{1}{x+1}+\dfrac{1}{y+2}+\dfrac{1}{z+3}\le1\)
Tìm GTNN của \(P=x+y+z+\dfrac{1}{x+y+z}\)
2, Cho các số x,y dương thỏa mãn đk: xy+yz+zx =671
CMR: \(\dfrac{x}{x^2-yz+2013}+\dfrac{y}{y^2-zx+2013}+\dfrac{z}{z^2-xy+2013}\ge\dfrac{1}{x+y+z}\)
2) \(\sum\dfrac{x}{x^2-yz+2013}=\sum\dfrac{x^2}{x^3-xyz+2013x}\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\dfrac{1}{x+y+z}\left(đpcm\right)\)