Now we give the proof of Theorem 1.
Let \(f=\rho^{\gamma }\varphi \), \(\rho =\sin \theta \), \(\gamma =- \frac{N-p}{p}\), by the calculation that appeared in [13], one has
$$\begin{aligned} \vert \nabla_{S}f\vert ^{p} =&\bigl\vert \nabla_{S}\bigl(\rho^{\gamma }\bigr)\varphi +\rho^{\gamma} \nabla_{S}\varphi \bigr\vert ^{p} \\ \geq& \bigl\vert \nabla_{S}\bigl(\rho^{\gamma }\bigr)\varphi \bigr\vert ^{p}+p\bigl\vert \nabla_{S}\bigl( \rho^{\gamma }\bigr)\varphi \bigr\vert ^{p-2}\vert \nabla_{S}\bigl(\rho^{\gamma }\bigr)\varphi \cdot \rho^{\gamma }\nabla_{S}\varphi, \end{aligned}$$
then integration by parts gives
$$\begin{aligned} \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV \geq \vert \gamma \vert ^{p} \int_{S^{N}} \rho^{\gamma p-p}\vert \nabla_{S}\rho \vert ^{p}\vert \varphi \vert ^{p}\,dV-\frac{\vert \gamma \vert ^{p-2}\gamma }{\gamma p-p+2} \int_{S^{N}}\Delta_{S} \bigl( \rho^{\gamma p-p+2} \bigr) \vert \varphi \vert ^{p}\,dV. \end{aligned}$$
Since
$$\vert \nabla_{S}\rho \vert =\vert \nabla_{S}\sin \theta \vert =\vert \cos \theta \vert $$
and
$$\Delta_{S} \bigl( \rho^{\gamma p-p+2} \bigr) =\Delta_{S} \bigl( \sin ^{2-N} \theta \bigr) =(N-2)\sin ^{2-N}\theta, $$
one has
$$\begin{aligned} \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV \geq \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}}\frac{\vert \cos \theta \vert ^{p}\vert \varphi \vert ^{p}}{\sin ^{N}\theta }\,dV- \biggl( \frac{N-p}{p} \biggr) ^{p-1} \int_{S^{N}}\frac{\vert \varphi \vert ^{p}}{\sin ^{N-2}\theta }\,dV. \end{aligned}$$
While
$$\vert \cos \theta \vert ^{p}=\bigl\vert \cos ^{2}\theta \bigr\vert ^{\frac{p}{2}}= \bigl( 1-\sin ^{2}\theta \bigr) ^{\frac{p}{2}}\geq 1-\frac{p}{2}\sin ^{2}\theta\quad \mbox{for } p \geq 2 $$
and
$$\vert \cos \theta \vert ^{p}\geq \cos ^{2}\theta \quad \mbox{for } 1< p\leq 2. $$
The previous inequality can be written as follows:
$$\begin{aligned}& \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV \\& \quad \geq \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}}\frac{\vert \varphi \vert ^{p}}{\sin ^{N}\theta }\,dV \\& \qquad {}- \biggl( \biggl( \frac{N-p}{p} \biggr) ^{p-1} +\min \biggl\{ \frac{p}{2},1 \biggr\} \biggl( \frac{N-p}{p} \biggr) ^{p} \biggr) \int_{S^{N}}\frac{\vert \varphi \vert ^{p}}{\sin ^{N-2}\theta }\,dV. \end{aligned}$$
Define \(C(N,p)= ( \frac{N-p}{p} ) ^{p-1} +\min \{ \frac{p}{2},1 \} ( \frac{N-p}{p} ) ^{p}\), then
$$\begin{aligned}& C(N,p) \int_{S^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p-2}\theta }\,dV+ \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV \\& \quad \geq \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p}\theta }\,dV. \end{aligned}$$
(4)
In order to get our result, we rewrite the above inequality as
$$\begin{aligned}& C(N,p) \int_{S^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p-2}\theta }\,dV+ \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV \\& \quad \geq \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}} \biggl( \frac{\vert f\vert ^{p}}{ \theta^{p}}+\frac{\vert f\vert ^{p}}{(\pi -\theta)^{p}} \biggr)\,dV \\& \qquad {}+ \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p-2} \theta } \biggl( \frac{1}{\sin ^{2}\theta }- \frac{\sin ^{p-2}\theta }{\theta ^{p}}-\frac{\sin ^{p-2}\theta }{(\pi -\theta)^{p}} \biggr)\,dV. \end{aligned}$$
(5)
As we know, for \(p>2\),
$$\begin{aligned} \lim_{\theta \rightarrow 0^{+}} \biggl( \frac{1}{\sin ^{2}\theta }-\frac{\sin ^{p-2}\theta }{\theta^{p}}- \frac{\sin ^{p-2}\theta }{(\pi -\theta)^{p}} \biggr) =&\lim_{\theta \rightarrow 0^{+}}\frac{\theta^{p}-\sin ^{p}\theta }{ \theta^{p+2}} \\ =&\lim _{\theta \rightarrow 0^{+}}\frac{1- ( \frac{\sin \theta }{\theta } ) ^{p}}{\theta^{2}} \\ \geq& \lim_{\theta \rightarrow 0^{+}}\frac{1- ( \frac{\sin \theta }{\theta } ) ^{2}}{\theta^{2}} \\ =&\frac{1}{3}>0, \end{aligned}$$
and for \(p=2\),
$$\lim_{\theta \rightarrow 0^{+}} \biggl( \frac{1}{\sin ^{2}\theta }-\frac{\sin ^{p-2}\theta }{\theta^{p}}- \frac{\sin ^{p-2}\theta }{(\pi -\theta)^{p}} \biggr) = \frac{1}{3}-\frac{1}{ \pi }>0. $$
Also, by a similar calculation, one has for \(p>2\)
$$\lim_{\theta \rightarrow \pi^{-}} \biggl( \frac{1}{\sin ^{2}\theta }-\frac{\sin ^{p-2}\theta }{\theta^{p}}- \frac{\sin ^{p-2}\theta }{(\pi - \theta)^{p}} \biggr) \geq \frac{1}{3}>0, $$
and for \(p=2\)
$$\lim_{\theta \rightarrow \pi^{-}} \biggl( \frac{1}{\sin ^{2}\theta }-\frac{\sin ^{p-2}\theta }{\theta^{p}}- \frac{\sin ^{p-2}\theta }{(\pi - \theta)^{p}} \biggr) \geq \frac{1}{3}-\frac{1}{\pi }>0. $$
Therefore, there exist a constant \(\theta_{1}>0\) small enough and \(\theta_{2}<\pi \) close to π such that for any \(\theta \in (0, \theta_{1}]\cup [\theta_{2},\pi)\), one has that
$$\frac{1}{\sin ^{2}\theta }-\frac{\sin ^{p-2}\theta }{\theta^{p}}-\frac{\sin ^{p-2}\theta }{(\pi -\theta)^{p}}>0 \quad \mbox{for } p\geq 2. $$
Thus from inequality (5), one has
$$\begin{aligned}& C(N,p) \int_{S^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p-2}\theta }\,dV+ \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV \\& \quad \geq \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}} \biggl( \frac{\vert f\vert ^{p}}{ \theta^{p}}+\frac{\vert f\vert ^{p}}{(\pi -\theta)^{p}} \biggr)\,dV \\& \qquad {}+ \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}\cap ( [0,\theta_{1}]\cup [\theta_{2},\pi ] ) }\frac{\vert f\vert ^{p}}{\sin ^{p-2}\theta } \biggl( \frac{1}{\sin ^{2}\theta }- \frac{\sin ^{p-2}\theta }{ \theta^{p}}-\frac{\sin ^{p-2}\theta }{(\pi -\theta)^{p}} \biggr)\,dV \\& \qquad {}+ \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}\cap [\theta_{1},\theta_{2}]}\frac{\vert f\vert ^{p}}{\sin ^{p-2} \theta } \biggl( \frac{1}{\sin ^{2}\theta }- \frac{\sin ^{p-2}\theta }{\theta ^{p}}-\frac{\sin ^{p-2}\theta }{(\pi -\theta)^{p}} \biggr)\,dV \\& \quad \geq \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}} \biggl( \frac{\vert f\vert ^{p}}{ \theta^{p}}+\frac{\vert f\vert ^{p}}{(\pi -\theta)^{p}} \biggr)\,dV-C_{1} \int_{S ^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p-2}\theta }\,dV, \end{aligned}$$
where \(C_{1}=\sup_{\theta \in [\theta_{1},\theta_{2}]} ( \frac{1}{\sin ^{2}\theta }-\frac{\sin ^{p-2}\theta }{\theta^{p}}-\frac{\sin ^{p-2} \theta }{(\pi -\theta)^{p}} ) \).
Let \(C=C(N,p)+C_{1}\), one has that for \(p\geq 2 \)
$$\begin{aligned} C \int_{S^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p-2}\theta }\,dV+ \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV \geq \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}} \biggl( \frac{\vert f\vert ^{p}}{ \theta^{p}}+\frac{\vert f\vert ^{p}}{(\pi -\theta)^{p}} \biggr)\,dV , \end{aligned}$$
which is exactly inequality (2).
While for \(1< p<2\), we get from inequality (4) that
$$\begin{aligned}& C(N,p) \int_{S^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p-2}\theta }\,dV+ \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV \\& \quad \geq \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}} \biggl( \frac{\vert f\vert ^{p}}{ \theta^{p}}+\frac{\vert f\vert ^{p}}{(\pi -\theta)^{p}} \biggr)\,dV \\& \qquad {}+ \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}}\vert f\vert ^{p} \biggl( \frac{1}{\sin ^{p}\theta }-\frac{1}{\theta^{p}}-\frac{1}{(\pi -\theta)^{p}} \biggr)\,dV. \end{aligned}$$
(6)
By a similar calculation, we get that for \(1< p<2\)
$$\begin{aligned} -\frac{1}{(\pi -\theta)^{p}} \leq& \biggl( \frac{1}{\sin ^{p}\theta }-\frac{1}{ \theta^{p}}- \frac{1}{(\pi -\theta)^{p}} \biggr) \\ =&\frac{1- ( \frac{\sin \theta }{\theta } ) ^{p}}{\sin ^{p}\theta }- \frac{1}{(\pi -\theta)^{p}}\leq \frac{1- ( \frac{\sin \theta }{ \theta } ) ^{2}}{\sin ^{2}\theta } -\frac{1}{(\pi -\theta)^{p}}. \end{aligned}$$
Since
$$\lim_{\theta \rightarrow 0^{+}}-\frac{1}{(\pi -\theta)^{p}}=-\frac{1}{ \pi^{p}},\qquad \lim _{\theta \rightarrow 0^{+}}\frac{1- ( \frac{\sin \theta }{\theta } ) ^{2}}{\sin ^{2}\theta } -\frac{1}{(\pi -\theta)^{p}}=\frac{1}{3}- \frac{1}{\pi^{p}}, $$
we get that there exists a constant \(\theta_{1}>0\) small enough such that
$$\sup_{\theta \in [0,\theta_{1}]}\biggl\vert \biggl( \frac{1}{\sin ^{p} \theta }- \frac{1}{\theta^{p}}-\frac{1}{(\pi -\theta)^{p}} \biggr) \biggr\vert < + \infty. $$
Furthermore, by a similar calculation, we get that there exists a constant \(\theta_{2}\) close to π such that
$$\sup_{\theta \in [\theta_{2},\pi ]}\biggl\vert \biggl( \frac{1}{\sin ^{p}\theta }- \frac{1}{\theta^{p}}-\frac{1}{(\pi -\theta)^{p}} \biggr) \biggr\vert < + \infty. $$
Therefore, we get that
$$C_{2}:=\sup_{\theta \in [0,\pi ]}\biggl\vert \biggl( \frac{1}{\sin ^{p} \theta }-\frac{1}{\theta^{p}}-\frac{1}{(\pi -\theta)^{p}} \biggr) \biggr\vert < + \infty. $$
Then, by inequality (6), one has
$$\begin{aligned}& C(N,p) \int_{S^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p-2}\theta }\,dV+ \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV \\& \quad \geq \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}} \biggl( \frac{\vert f\vert ^{p}}{ \theta^{p}}+\frac{\vert f\vert ^{p}}{(\pi -\theta)^{p}} \biggr)\,dV -C_{2} \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}}\vert f\vert ^{p}\,dV. \end{aligned}$$
Let \(C=C(N,p)+C_{2} ( \frac{N-p}{p} ) ^{p}\), we get for \(1< p<2\)
$$\begin{aligned} C \int_{S^{N}}\vert f\vert ^{p}\,dV+ \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV \geq \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{S^{N}} \biggl( \frac{\vert f\vert ^{p}}{\theta^{p}}+\frac{\vert f\vert ^{p}}{( \pi -\theta)^{p}} \biggr)\,dV, \end{aligned}$$
which is exactly inequality (3).
Now we prove that \(( \frac{N-p}{p} ) ^{p}\) is the best constant of inequalities (2) and (3).
Let \(\varphi (t)\in [0,1]\) be a cut-off function such that \(\varphi (t) \equiv 1\) for \(\vert t\vert \leq 1\); \(\varphi (t)\equiv 0\) for \(\vert t\vert >2\). Define \(H(t)=1-\varphi (t)\) and
$$f_{\epsilon }(\theta)=H \biggl( \frac{\theta }{\epsilon } \biggr) \theta^{\frac{p-N}{p}}\quad \mbox{for } 0< \theta \leq \pi;\qquad f_{ \epsilon }(\theta)=0 \quad \mbox{for } \theta =0. $$
Then we have
$$\begin{aligned}& \int_{S^{N}}\frac{\vert f_{\epsilon }\vert ^{p}}{\sin ^{p-2}\theta }\,dV=\bigl\vert S^{N-1}\bigr\vert \int_{\epsilon }^{\pi }H^{p} \biggl( \frac{\theta }{\epsilon } \biggr) \theta^{p-N}\sin ^{N-p+1}\theta \,d\theta \leq \bigl\vert S^{N-1}\bigr\vert \frac{\pi^{2}- \epsilon^{2}}{2}, \end{aligned}$$
(7)
$$\begin{aligned}& \int_{S^{N}}\vert f_{\epsilon }\vert ^{p}\,dV=\bigl\vert S^{N-1}\bigr\vert \int_{\epsilon }^{\pi }H ^{p} \biggl( \frac{\theta }{\epsilon } \biggr) \theta^{p-N}\sin ^{N-1}\theta \,d\theta \leq \bigl\vert S^{N-1}\bigr\vert \frac{\pi^{p}-\epsilon^{p}}{p}, \end{aligned}$$
(8)
$$\begin{aligned}& \int_{S^{N}}\frac{\vert f_{\epsilon }\vert ^{p}}{\theta^{p}}\,dV=\bigl\vert S^{N-1}\bigr\vert \int_{\epsilon }^{\pi }H^{p} \biggl( \frac{\theta }{\epsilon } \biggr) \theta^{-N}\sin ^{N-1}\theta \,d\theta \geq \bigl\vert S^{N-1}\bigr\vert \int_{2\epsilon } ^{\pi }\theta^{-N}\sin ^{N-1} \theta \,d\theta, \end{aligned}$$
(9)
$$\begin{aligned}& \int_{S^{N}}\vert \nabla_{S}f_{\epsilon }\vert ^{p}\,dV \\& \quad =\bigl\vert S^{N-1}\bigr\vert \int_{\epsilon }^{\pi }\biggl\vert \frac{1}{\epsilon }H^{\prime} \biggl( \frac{\theta }{\epsilon } \biggr) \theta^{\frac{p-N}{p}} + \frac{p-N}{q}H \biggl( \frac{\theta }{\epsilon } \biggr) \theta^{- \frac{N}{p}}\biggr\vert ^{p}\sin ^{N-1}\theta \,d\theta \\& \quad =\bigl\vert S^{N-1}\bigr\vert \int_{\epsilon }^{2\epsilon }\biggl\vert \frac{1}{\epsilon }H ^{\prime} \biggl( \frac{\theta }{\epsilon } \biggr) \theta^{\frac{p-N}{p}} + \frac{p-N}{p}H \biggl( \frac{\theta }{\epsilon } \biggr) \theta^{- \frac{N}{p}}\biggr\vert ^{p}\sin ^{N-1}\theta \,d\theta \\& \qquad {}+\bigl\vert S^{N-1}\bigr\vert \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{2\epsilon }^{\pi } \biggl\vert H \biggl( \frac{\theta }{\epsilon } \biggr) \theta^{-\frac{N}{p}}\biggr\vert ^{p}\sin ^{N-1} \theta \,d\theta \\& \quad \leq C_{1}\max_{t\in [0,2]}\bigl\vert H^{\prime}(t)\bigr\vert +C_{2} \int_{\epsilon } ^{2\epsilon }\theta^{-1}\,d\theta +\bigl\vert S^{N-1}\bigr\vert \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{2\epsilon }^{\pi }\theta^{-N}\sin ^{N-1} \theta \,d\theta \\& \quad \leq C_{4}+\bigl\vert S^{N-1}\bigr\vert \biggl( \frac{N-p}{p} \biggr) ^{p} \int_{2\epsilon } ^{\pi }\theta^{-N}\sin ^{N-1} \theta \,d\theta. \end{aligned}$$
(10)
Therefore, for \(p\geq 2\), from inequalities (7), (9) and (10), one has
$$\begin{aligned}& \inf_{f\in C^{\infty }(S^{N})}\frac{C\int_{S^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p-2}\theta }\,dV+\int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV}{\int_{S^{N}}\frac{\vert f\vert ^{p}}{ \theta^{p}}\,dV} \\& \quad \leq \frac{C\int_{S^{N}}\frac{\vert f_{\epsilon }\vert ^{p}}{\sin ^{p-2}\theta }\,dV+\int_{S^{N}}\vert \nabla_{S_{\epsilon }}f\vert ^{p}\,dV}{\int_{S^{N}} \frac{\vert f_{\epsilon }\vert ^{p}}{\theta^{p}}\,dV} \\& \quad \leq \frac{\vert S^{N-1}\vert \frac{\pi^{2}-\epsilon^{2}}{2}+ C_{4}+\vert S^{N-1}\vert ( \frac{N-p}{p} ) ^{p}\int_{2\epsilon }^{\pi }\theta^{-N}\sin ^{N-1}\theta \,d\theta }{\vert S^{N-1}\vert \int_{2\epsilon }^{\pi }\theta^{-N}\sin ^{N-1}\theta \,d\theta }, \end{aligned}$$
then passing to the limit as \(\epsilon \rightarrow 0^{+}\), we have
$$ \inf_{f\in C^{\infty }(S^{N})}\frac{C\int_{S^{N}}\frac{\vert f\vert ^{p}}{\sin ^{p-2}\theta }\,dV+\int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV}{\int_{S^{N}}\frac{\vert f\vert ^{p}}{ \theta^{p}}\,dV}\leq \biggl( \frac{N-p}{p} \biggr) ^{p}. $$
(11)
Since \(\lim_{\epsilon \rightarrow 0^{+}}\int_{2\epsilon }^{ \pi }\theta^{-N}\sin ^{N-1}\theta \,d\theta \rightarrow +\infty \).
While for \(1< p<2\), from (8), (9) and (10), one has
$$\begin{aligned}& \inf_{f\in C^{\infty }(S^{N})}\frac{C\int_{S^{N}}\vert f\vert ^{p}\,dV+ \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV}{\int_{S^{N}}\frac{\vert f\vert ^{p}}{\theta ^{p}}\,dV} \\& \quad \leq \frac{C\int_{S^{N}}\vert f_{\epsilon }\vert ^{p}\,dV+\int_{S^{N}}\vert \nabla_{S_{\epsilon }}f\vert ^{p}\,dV}{\int_{S^{N}}\frac{\vert f_{\epsilon }\vert ^{p}}{ \theta^{p}}\,dV} \\& \quad \leq \frac{\vert S^{N-1}\vert \frac{\pi^{p}-\epsilon^{p}}{p}+ C_{4}+\vert S^{N-1}\vert ( \frac{N-p}{p} ) ^{p}\int_{2\epsilon }^{\pi }\theta^{-N}\sin ^{N-1}\theta \,d\theta }{\vert S^{N-1}\vert \int_{2\epsilon }^{\pi }\theta^{-N}\sin ^{N-1}\theta \,d\theta }, \end{aligned}$$
then passing to the limit as \(\epsilon \rightarrow 0^{+}\), we have
$$ \inf_{f\in C^{\infty }(S^{N})}\frac{C\int_{S^{N}}\vert f\vert ^{p}\,dV+ \int_{S^{N}}\vert \nabla_{S}f\vert ^{p}\,dV}{\int_{S^{N}}\frac{\vert f\vert ^{p}}{\theta ^{p}}\,dV}\leq \biggl( \frac{N-p}{p} \biggr) ^{p}. $$
(12)
Therefore, from (2), (3), (11) and (12), we get that \(( \frac{N-p}{p} ) ^{p}\) is the best constant of inequalities (2) and (3). Proof of Theorem 1 is finished.