Cho a,b,c>0 thỏa mãn \(\left(ab\right)^2+\left(bc\right)^2+\left(ac\right)^2\ge\left(abc\right)^2\)

Chứng minh rằng \(\frac{\left(ab\right)^2}{\left(a^2+b^2\right)c^3}+\frac{\left(bc\right)^2}{\left(b^2+c^2\right)a^3}+\frac{\left(ac\right)^2}{\left(a^2+c^2\right)b^3}\ge\frac{\sqrt{3}}{2}\)

cho 3 số dương x,y,z thỏa mãn xyz=1 và \(x+y+z>\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\).

Chứng minh rằng trong 3 số x,y,z chỉ có đúng 1 số lớn hơn 1

\(\sqrt[3]{x^2+4}=\sqrt{x-1}+2x-3\)

\(\left\{{}\begin{matrix}x+y=zt\\z+t=xy\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x^2-6x-2y=15\\\left(x^2-3x\right)y=-2\left(z+3\right)\\x^2y^2+2y+12\le4z\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x-y+z=2\\2x^2-xy+x-2z=1\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{2x^2}{x^2+1}=y\\\dfrac{3y^3}{y^4+y^2+y}=z\\\dfrac{4z^4}{z^6+z^4+z^2+1}=x\end{matrix}\right.\)

\(\left\{{}\begin{matrix}x^2-y^2+xy=1\\3x+y=y^2+3\end{matrix}\right.\)