Prove that $\varphi(m)+ \tau(m)\leqslant m+1$ 
Prove that $\varphi(m)+ \tau(m)\leqslant m+1$ where $m\in \mathbb N$

I wrote $m:=p_1^{\alpha_ 1}....p_s^{\alpha _s}$
$$\varphi(m)=p_1^{\alpha_1}(p_1-1)...p_s^{\alpha_s}(p_s-1)$$
$$\tau(m)=(\alpha_1+1)...(\alpha_s+1)$$
Then $\varphi(m)+ \tau(m)=\prod\limits_{i=1}^{s}p_i^{\alpha_i-1}(p_i-1)+\sum\limits_{i=1}^{s}(\alpha_i+1)\overset{?}{\leqslant}\prod\limits_{i=1}^{s} p_i^{\alpha_i}+1$
and I'm stuck here, how can I prove that $\prod\limits_{i=1}^{s}p_i^{\alpha_i-1}(p_1-1)+\sum\limits_{i=1}^{s}(\alpha_i+1)\overset{?}{\leqslant}\prod\limits_{i=1}^{s} p_i^{\alpha_i}+1$
or maybe is there an easy way to prove that $\varphi(m)+ \tau(m)\leqslant m+1$ 

Fix $m\in\mathbb{N}$ and let $A=\{a\in\mathbb{N}:a\leq m,\gcd{(a,m)}=1\},$ $B=\{a\in\mathbb{N}:a\mid m\}$. Now $\lvert A \rvert = \varphi(m)$ and $\lvert B \rvert = \tau(m)$. Hence $$\varphi(m)+\tau(m) = \lvert A \rvert + \lvert B \rvert = \lvert A \cup B \rvert + \lvert A\cap B \rvert.$$
But now note that $A\cup B\subseteq\{1,2,\ldots,m\}$ and $A\cap B = \{1\}$, so $\lvert A\cup B \rvert \leq m$ and $\lvert A\cap B \rvert = 1$. The result follows.
