Các câu hỏi tương tự
cho x y z > 0 và x+y+z=1. Tìm GTNN của \(P=\sqrt{\frac{xy}{z+xy}}+\sqrt{\frac{yz}{x+yz}}+\sqrt{\frac{zx}{y+zx}}\)
cho x+y+z=1. CMR \(\sqrt{\frac{xy}{xy+z}}+\sqrt{\frac{yz}{yz+x}}+\sqrt{\frac{zx}{zx+y}}< =\frac{3}{2}\)giúp mình với!!!!!!!!!
Cho ba số x, y, z>0. Chứng minh rằng:
\(\frac{2}{x+\sqrt{yz}}+\frac{2}{y+\sqrt{zx}}+\frac{2}{z+\sqrt{xy}}\le\frac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{\sqrt{xyz}}\)
Cho x,y,z >0 tm xy+yz+zx=xyz. Tìm GTLN của:
\(A=\frac{1}{\sqrt{x^2-xy+y^2}}+\frac{1}{\sqrt{y^2-yz+z^2}}+\frac{1}{\sqrt{z^2-zx+x^2}}\)
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 \(\hept{\begin{cases}x,y,z>0\\xy+yz+zx=1\end{cases}}\). Chứng minh rằng:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge3+\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{x^2}}+\sqrt{\frac{\left(y+z\right)\left(y+x\right)}{y^2}}+\sqrt{\frac{\left(z+x\right)\left(z+y\right)}{z^2}}\)
cho \(x,y,z>0;xy+yz+zx=1\)
chứng minh
\(\frac{x}{\sqrt{x^2+1}}+\frac{y}{\sqrt{y^2+1}}+\frac{z}{\sqrt{z^2+1}}\le\frac{3}{2}\)
cho x,y,z>0 thoả mãn x2+y2+z2=3. Chứng minh rằng:
\(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\ge xy+yz+zx\)
CHO X,Y,Z LÀ CÁC SỐ THỰC KHÔNG ÂM THỎA MÃN X+Y+Z=3 VÀ XY+YZ+ZX KHÁC 0 . CMR :
\(\frac{x+1}{y+1}+\frac{y+1}{z+1}+\frac{z+1}{x+1}\le\frac{25}{3\sqrt[3]{4.\left(xy+yz+zx\right)}}\)
thanks in advance <3