(x+y+z)^2/3>=(x^2+y^2+z^2+2xy+2yz+2zx)/3>=3(xy+yz+zx)/3=xy+yz+zx(do x^2+y^2+z^2>=xy+yz+zx)(1)
(xy+yz)/2>=y√xz;(yz+zx)/2>=z√xy;(zx+xy)/2>=x√yz(BĐT Cô-si)
Cộng theo vế >>>xy+yz+zx>=y√xz+z√xy+x√yz(2)
Từ(1),(2) >>>đpcm
Cho x;y;z > 0 thỏa mãn x2 + y2 + z2 = 3
CMR: \(\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\ge xy+yz+zx\)
Cho các số thực dương x, y, z thỏa mãn \(x^2+y^2+z^2=3\)
\(CMR:\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}\ge xy+yz+xz\)
Cho x,y,z>0. CMR:
\(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)\)
cho \(x;y;z>0\)
\(xy+yz+xz=xyz\)
và \(\left(x+y\right)\left(\frac{1}{z}+\frac{1}{xy}\right)+\left(y+z\right)\left(\frac{1}{x}+\frac{1}{yz}\right)+\left(x+z\right)\left(\frac{1}{y}+\frac{1}{xz}\right)=1\)
tính giá trị của biểu thức
\(A=\sqrt{\frac{\left(2x+yz\right)\left(2y+xz\right)}{\left(y+z\right)\left(x+z\right)}}+\sqrt{\frac{\left(2y+xz\right)\left(2z+xy\right)}{\left(x+z\right)\left(x+y\right)}}+\sqrt{\frac{\left(2z+xy\right)\left(2x+yz\right)}{\left(x+y\right)\left(y+z\right)}}\)
Cho các số dương x,y,z thỏa mãn x2+y2+z2=1.
Chứng minh: \(\frac{x+y+z}{xy+yz+xz}\ge\sqrt{3}+\frac{1}{2\sqrt{3}}[\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2]\)
tìm Max của\(P=\frac{x}{\sqrt{yz\left(1+x^2\right)}}+\frac{y}{\sqrt{zx\left(1+y^2\right)}}+\frac{z}{\sqrt{xy\left(1+z^2\right)}}\)với x y z > 0 và xy+yz+xz=xyz
Cho x, y, z >0 thỏa x + y + z >= 3. Chứng minh rằng : \(\frac{x^2}{x+\sqrt{yz}}+\frac{y^2}{y+\sqrt{xz}}+\frac{z^2}{z+\sqrt{xy}}\ge\frac{3}{2}\)
đặt \(A=\frac{\sqrt{yz}}{x+3\sqrt{yz}}+\frac{\sqrt{zx}}{y+3\sqrt{zx}}+\frac{\sqrt{xy}}{z+3\sqrt{xy}}\)
\(\Rightarrow1-3A=\frac{x}{x+3\sqrt{yz}}+\frac{y}{y+3\sqrt{zx}}+\frac{z}{z+3\sqrt{xy}}\)
\(\ge\frac{x}{x+\frac{3}{2}\left(y+z\right)}+\frac{y}{y+\frac{3}{2}\left(z+x\right)}+\frac{z}{z+\frac{3}{2}\left(x+y\right)}\)
\(=\frac{2x}{2x+3\left(y+z\right)}+\frac{2y}{2y+3\left(z+x\right)}+\frac{2z}{2z+3\left(x+y\right)}\)
\(=\frac{2x^2}{2x^2+3xy+3xz}+\frac{2y^2}{2y^2+3yz+3xy}+\frac{2z^2}{2z^2+3zx+3yz}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+6\left(xy+yz+zx\right)}=\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+2\left(xy+yz+zx\right)}\)
\(\ge\frac{2\left(x+y+z\right)^2}{2\left(x+y+z\right)^2+\frac{2}{3}\left(x+y+z\right)^2}=\frac{2\left(x+y+z\right)^2}{\frac{8}{3}\left(x+y+z\right)^2}=\frac{3}{4}\)
\(\Rightarrow1-3A\ge\frac{3}{4}\Rightarrow A\le\frac{3}{4}\left(Q.E.D\right)\)
CMR: \(\sqrt{x^2+xy+y^2}+\sqrt{y^2+yz+z^2}+\sqrt{z^2+zx+x^2}\ge\sqrt{3}\left(x+y+z\right)\)
Với x;y;z > 0.
