I know that $u\in B(H)$ is a normal operator if $uu^*=u^*u$.
I know that if $u$ is subnormal then $uu^*\neq u^*u$ (like unilateral shift operator).
My problem is:"If $u\in B(H)$ is subnormal then $u^*u\geq uu^* $"
Yes, it is. Condition $u^*u\ge uu^*$ is called hyponormality.
Proposition: Let $M$ a subspace of $H$ and $T\in B(H)$. If $M$ is $T$-invariant, then $(T|_M)^*=PT^*|_M$, where $P$ is the orthogonal projection onto $M$.
Now, if $S$ is subnormal in $H$, there is a superspace $K\ge H$ and a normal operator $N\in B(K)$ such that $N|_H=S$ and $H$ is $N$-invariant. Let $P$ the orthogonal projection of $K$ onto $H$. By the proposition, S^=PN^|_H and $\|S^*x\|^2\le\|Nx\|^2$ for all $x\in H$. Thus $\langle S^*x,S^*x\rangle\le\langle Nx,Nx\rangle$ for all $x\in H$, and, since $N$ is an extension of $S$ in $H$, this is equivalent to $\langle SS^*x,x\rangle\le\langle Sx,Sx\rangle=\langle S^*Sx,x\rangle$ for all $x\in H$, which proves $SS^*\le S^*S$.
On the other hand, no. Normal extensions aren't unitary equivalent, but minimal normal extension always exists, and any to minimal normal extensions are unitary equivalent.