Các câu hỏi tương tự
31 tháng 12 2020 lúc 21:59
Cho x,y,z là các số thực thỏa mãn x+y+z = 0
Chứng minh \(P=\frac{x\left(x+2\right)}{2x^2+1}+\frac{y\left(y+2\right)}{2y^2+1}+\frac{z\left(z+2\right)}{2z^2+1}\ge0\)
Cho x>0,y>0,z>0, xyz=1
Tìm GTNN
\(P=\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(x+z\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}.\)
Cho x>0, y>0,z>0,xyz=1. CMR \(P=\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}\) lớn hơn hoặc bằng 2
Cho x,y,z>0 thỏa mãn xyz=1. Tìm min \(P=\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}\)
Cho x,y,z>0 va xyz=1. Tim Min cua \(P=\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}\)
chị QAta có đề bài frac{x^2}{y}-2x+y+frac{y^2}{z}-2y+z+frac{z^2}{x}-2z+x+left(x+y+zright)-left(x-yright)^2-left(y-zright)^2-left(z-xright)^2frac{left(x-yright)^2}{y}-left(x-yright)^2+...+left(x+y+zright)left(x-yright)^2left(frac{1}{y}-1right)+....+left(x+y+zright)mà sqrt{x}+sqrt{y}+sqrt{z}1Rightarrow x,y,zinleft[0;1right] frac{1}{y}-y0 Age x+y+zgefrac{left(sqrt{x}+sqrt{y}+sqrt{z}right)^2}{3}frac{1}{3}
Đọc tiếp
chị QA
ta có đề bài <=>
\(\frac{x^2}{y}-2x+y+\frac{y^2}{z}-2y+z+\frac{z^2}{x}-2z+x+\left(x+y+z\right)-\left(x-y\right)^2-\left(y-z\right)^2-\left(z-x\right)^2\)
=\(\frac{\left(x-y\right)^2}{y}-\left(x-y\right)^2+...+\left(x+y+z\right)\)
=\(\left(x-y\right)^2\left(\frac{1}{y}-1\right)+....+\left(x+y+z\right)\)
mà \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\Rightarrow x,y,z\in\left[0;1\right]\)
=> \(\frac{1}{y}-y>0\)
=> \(A\ge x+y+z\ge\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}=\frac{1}{3}\)
Cho x,y,z là 3 số thực dương thỏa mãn xyz=1.Chứng minh bất đẳng thức
\(\frac{1}{\left(2x+y+z\right)^2}+\frac{1}{\left(x+2y+z\right)^2}+\frac{1}{\left(x+y+2z\right)^2}\le\frac{3}{16}\)
đặt Afrac{sqrt{yz}}{x+3sqrt{yz}}+frac{sqrt{zx}}{y+3sqrt{zx}}+frac{sqrt{xy}}{z+3sqrt{xy}}Rightarrow1-3Afrac{x}{x+3sqrt{yz}}+frac{y}{y+3sqrt{zx}}+frac{z}{z+3sqrt{xy}}gefrac{x}{x+frac{3}{2}left(y+zright)}+frac{y}{y+frac{3}{2}left(z+xright)}+frac{z}{z+frac{3}{2}left(x+yright)}frac{2x}{2x+3left(y+zright)}+frac{2y}{2y+3left(z+xright)}+frac{2z}{2z+3left(x+yright)}frac{2x^2}{2x^2+3xy+3xz}+frac{2y^2}{2y^2+3yz+3xy}+frac{2z^2}{2z^2+3zx+3yz}gefrac{2left(x+y+zright)^2}{2left(x^2+y^2+z^2right)+6left(xy+yz+zxr...
Đọc tiếp
đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)
\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)
\(\ge\frac{x}{x+\frac{3}{2}\left(y+z\right)}+\frac{y}{y+\frac{3}{2}\left(z+x\right)}+\frac{z}{z+\frac{3}{2}\left(x+y\right)}\)
\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)
\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)
cho các số x,y,z dương thỏa mãn \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1\) tìm GTNN
P=\(\frac{y^2z^2}{x\left(y^2+z^2\right)}+\frac{z^2x^2}{y\left(z^2+x^2\right)}+\frac{x^2y^2}{z\left(x^2+y^2\right)}\)