1) cho x,y>0 và \(x^4+y^4=2\) CMR: \(\dfrac{x^2}{y}+\dfrac{y^2}{x}\ge2\)
2) cho x,y,z > 0 và \(x^2+y^2+z^2=3\) CMR: \(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge3\)
1) Cho x,y > 0 và \(x^4+y^4=2\) CMR \(\dfrac{x^2}{y}+\dfrac{y^2}{x}\ge2\)
2) Cho x,y,z > 0 và \(x^2+y^2+z^2=3\) CMR \(\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}\ge3\)
cho x,y,z>0 và x3+y3+z3=1.
CMR:\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\)
Ta có với x,y,z >0 thì:\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Bất đẳng thức Cô si ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{x\sqrt{1-x^2}}\ge2\\ \Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\Leftrightarrow\dfrac{x^2}{\sqrt{1-x^2}}\ge2x^3\)
Tương tự: \(\dfrac{y^2}{\sqrt{1-y^2}}\ge2y^3;\dfrac{z^2}{\sqrt{1-z^2}}\ge2z^3\)
Từ đó ta có:\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\left(dpcm\right)\)
Cho \(\left\{{}\begin{matrix}x;y;z>=0\\x+y+z=2\end{matrix}\right.\) CMR \(\dfrac{1}{x^2-xy+y^2}+\dfrac{1}{y^2-yz+z^2}+\dfrac{1}{z^2-xz+x^2}\ge3\)
Không mất tính tổng quát, giả sử \(x\ge y\ge z\)
\(y^2-yz+z^2=y^2+\left(z-y\right)y\le y^2\Rightarrow\dfrac{1}{y^2-yz+z^2}\ge\dfrac{1}{y^2}\)
Tương tự: \(\dfrac{1}{z^2-xz+x^2}\ge\dfrac{1}{x^2}\)
\(\Rightarrow P\ge\dfrac{1}{x^2-xy+y^2}+\dfrac{1}{x^2}+\dfrac{1}{y^2}=\dfrac{1}{x^2-xy+y^2}+\dfrac{x^2-xy+y^2}{x^2y^2}+\dfrac{1}{xy}\)
\(P\ge2\sqrt{\dfrac{x^2-xy+y^2}{x^2y^2\left(x^2-xy+y^2\right)}}+\dfrac{1}{xy}=\dfrac{3}{xy}\ge\dfrac{12}{\left(x+y\right)^2}\ge\dfrac{12}{\left(x+y+z\right)^2}=3\)
Dấu "=" xảy ra khi \(\left(x;y;z\right)=\left(1;1;0\right)\) và hoán vị
CMR: Nếu \(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}\)=1 và\(\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\)=0 thì\(\dfrac{x^2}{y^2}+\dfrac{y^2}{z^2}+\dfrac{z^2}{x^2}\)=1
Cho x,y,z > 0 và \(x+y+z\le\dfrac{3}{2}\). CMR :
\(\sqrt{x^2+\dfrac{1}{x^2}}+\sqrt{y^2+\dfrac{1}{y^2}}+\sqrt{z^2+\dfrac{1}{z^2}}\ge\dfrac{3}{2}\sqrt{17}\)
Cho x;y;z>0 và không có 2 số nào đồng thời bằng 0.CMR:
\(\sqrt{\dfrac{x}{y+z}}+\sqrt{\dfrac{y}{z+x}}+\sqrt{\dfrac{z}{x+y}}\ge2\sqrt{1+\dfrac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}}\)
1) Rút gọn bt:
(x+y+z)3+(x-y-z)3+(y-x-z)3+(z-y-x)3
2)Tìm x,y,z t/m: 9x2+y2+2z2-18x+4z-6y+20=0
3)Cho \(\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}\)=1 và \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\)=0 . CMR:
\(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}\)=1
Cho x, y, z > 0 và xyz=1. CMR :
\(\dfrac{x^2}{1+y}+\dfrac{y^2}{1+z}+\dfrac{z^2}{1+z}\ge\dfrac{3}{2}\)
Đề sai nhé, \(\dfrac{z^2}{x+1}\) mới đúng nha
\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3}\left(\text{Svácxơ}\right)\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Ta có: \(x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow x+y+z+3\le2\left(x+y+z\right)\)
Cho x, y, z > 0 và \(x+y\le z\) . CMR :
\(\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)\ge\dfrac{27}{2}\)
\(VT=\left(x^2+y^2+z^2\right)\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=3+\dfrac{x^2+y^2}{z^2}+z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)+\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}\)
\(\dfrac{x^2}{y^2}+\dfrac{y^2}{x^2}>=2\cdot\sqrt{\dfrac{y^2}{x^2}\cdot\dfrac{x^2}{y^2}}=2\)
=>\(VT>=5+\left(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}\right)+\left(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}\right)+\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)\)
\(\dfrac{x^2}{z^2}+\dfrac{z^2}{16x^2}>=2\cdot\sqrt{\dfrac{x^2}{z^2}\cdot\dfrac{z^2}{16x^2}}=\dfrac{1}{2}\)
\(\dfrac{y^2}{z^2}+\dfrac{z^2}{16y^2}>=\dfrac{1}{2}\)
và \(\dfrac{1}{x^2}+\dfrac{1}{y^2}>=\dfrac{2}{xy}>=\dfrac{2}{\left(\dfrac{x+y}{2}\right)^2}=\dfrac{8}{\left(x+y\right)^2}\)
=>\(\dfrac{15}{16}z^2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}\right)>=\dfrac{15}{16}z^2\cdot\dfrac{8}{\left(x+y\right)^2}=\dfrac{15}{2}\left(\dfrac{z}{x+y}\right)^2=\dfrac{15}{2}\)
=>VT>=5+1/2+1/2+15/2=27/2