\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)
\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)
\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)
Cho x,y,z>0 /xyz=8.
Tìm min P= \(\dfrac{x^2}{\sqrt{\left(1+x^3\right)\left(1+y^3\right)}}+\dfrac{y^2}{\sqrt{\left(1+y^3\right)\left(1+z^3\right)}}+\dfrac{z^2}{\sqrt{\left(1+z^3\right)\left(1+x^3\right)}}\)
Tìm GTLN của biếu thức
P= x\(\sqrt{\dfrac{\left(1+y^2\right).\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+z^2\right).\left(1+x^2\right)}{1+y^2}}+z\sqrt{\dfrac{\left(1+x^2\right).\left(1+y^2\right)}{1+z^2}}\)
Cho x,y,z thỏa mãn xy+yz+xz = 1.Tính
\(S=x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
Cho 3 số dương x,y,z thõa mãn đk xy+yz+xz=1
Tính gt của bt:\(A=x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+x^2\right)\left(1+z^2\right)}{1+y^2}}+z\sqrt{\dfrac{\left(1+y^2\right)\left(1+x^2\right)}{1+z^2}}\)
Cho x, y, z dương thỏa mãn xyz = 1. Tìm GTLN:
P = \(\dfrac{1}{\left(3x+1\right)\left(y+z\right)+x}+\dfrac{1}{\left(3y+1\right)\left(z+x\right)+y}+\dfrac{1}{\left(3z+1\right)\left(x+y\right)+z}\)
Chứng minh rằng với mọi x, y, z > 0 ta có: \(\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\ge2+\dfrac{2\left(x+y+z\right)}{\sqrt[3]{xyz}}\)
Giải hệ phương trình sau : \(\left\{{}\begin{matrix}\sqrt{2x+1}+\sqrt{2y+1}=\dfrac{\left(x-y\right)^2}{2}\\\left(3x+2y\right)\left(y+1\right)=4-x^2\end{matrix}\right.\)
Cho 3 số dương x;y;z thỏa mãn x+y+z=6. CMR: \(x^2+y^2+z^2-xy-yz-xz+xyz\ge8\)
Cho x,y,z>0 và xy+yz+zx=1
Tính giá trị bt:
\(P=x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
cho x,y,z thỏa mãn \(x+y+z\le\dfrac{3}{2}\) . tìm GTNN của \(P=\dfrac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\dfrac{y\left(xz+1\right)^2}{y^2\left(xy+1\right)}+\dfrac{z\left(xy+1\right)^2}{x^2\left(yz+1\right)}\)
