Assume that $G+uv$ is Hamiltonian but $G$ is not. Then $G+uv$ has a Hamiltonian cycle containing the edge $uv$ Thus there exists a path $P=(x_1,\dots,x_n)$ in $G$ from $x_1=u$ to $x_n=v$ meeting all the vertices of $G$ If $x_i$ is adjacent to $x_1$ ($2\leq i\leq n$ then $x_{i-1}$ is not adjacent to $x_n$ for otherwise $(x_1,x_i,x_{i+1},\dots,x_n,x_{i-1},x_{i-2},\dots,x_1)$ is a Hamiltonian cycle of $G$ Thus $d(x_n)\leq (n-1)-d(x_1)$ that is $d(u)+d(v)\leq n-1$ a contradiction