easy to see,$\geq$ satifies add some poly. 1.if not inductionly prove $a_i=b_i,\forall i=1,2,...,n$ first $a_n=b_n$：consider coef. of $x^{n^2}$，in LHS $a_n^{n+1}+b_n^{n+1}$，in RHS $a_n^nb_n+a_nb_n^n$，LHS $\geq$ RHS,"$=$" iff $a_n=b_n$，so "$=$" must holds Assume $a_i=b_i,\forall i=k,k+1,...,n$ \[a_nf(x)^n+a_{n-1}f(x)^{n-1}+...+a_kf(x)^k+...+a_0+b_ng(x)^n+b_{n-1}g(x)^{n-1}+...+b_kg(x)^k+...+b_0\]\[\geq a_ng(x)^n+a_{n-1}g(x)^{n-1}+...+a_kg(x)^k+...+a_0+b_nf(x)^n+b_{n-1}f(x)^{n-1}+...+b_kf(x)^k+...+b_0\] i.e. \[a_{k-1}f(x)^{k-1}+...+a_0+b_{k-1}g(x)^{k-1}+...+b_0\geq a_{k-1}g(x)^{k-1}+...+a_0+b_{k-1}f(x)^{k-1}+...+b_0\] Easy to see $x^q$ have same coef. both side $\forall q>n(k-2)+k-1$(cause they choosen from $a_n$ to $a_k$)，consider coef. of $x^{n(k-2)+k-1}$，in LHS $(k-1)a_n^{k-2}(a_{k-1}^2+b_{k-1}^2)$，in RHS $(k-1)a_n^{k-2}\times 2a_{k-1}b_{k-1}$，AM-GM get LHS$\geq$RHS,"$=$" iff $a_{k-1}=b_{k-1}$，so "$=$" must holds get $f(x)=g(x)$ by induction ，contradiction！ 2.$f(x)\geq g(x)$ iff exists $M$ large s.t. $\forall m>M,f(m)\geq g(m)$ WLOG $f(x)\geq g(x)$，only need to prove $(f-g)(f(x))\geq (f-g)(g(x))$，and $f-g$ has positive leading coef.，then there are $M$ large s.t.$\forall m>M$：(1)$(f-g)(m)$ non-decreasing，(2)$f(m)\geq g(m)$，so $(f-g)(f(m))\geq (f-g)(g(m))$，done!