We write (here, $\mathcal F$ means the Fourier transform on $\mathcal S(\mathbb R^d)$) $$P_{\ge N}f= \mathcal F^{-1}((1-\varphi(\xi/2N))\mathcal F(f)(\xi))$$ for $N=2^m$ with $m\in\mathbb Z$ and a fixed non-negative radial smooth function $\varphi(\xi)$ on $\mathbb R^d$ which equals $1$ on $|\xi|\le 1$ and is supported on $|\xi|\le 2$.
Equation (A.2) of Tao’s Nonliear dispersive equations says: for any $s\ge0$, $1\le p\le\infty$ and $f\in\mathcal S(\mathbb R^d)$, we have the following inequality $$\| P_{\ge N}f \|_{L^p(\mathbb R^d)}\le C_{p,s,d} N^{-s}\| |\nabla|^sP_{\ge N}f\|_{L^p(\mathbb R^d)}$$ which I don’t know how to prove.
There are several variants of this inequality, for example, the low frequency version and versions with two different exponents $p$ and $q$. All of them can be proved easily by using $\varphi(\xi/2N) \varphi(\xi/N) = \varphi(\xi/N)$ and Young’s inequality for convolutions. However for the above one, we can’t apply Young’s inequality essentially because $1-\varphi(\xi/N)$ has infinity $L^1$-norm.
A professor in my department kindly taught me a way to show the case of $d=1$ (originally I thought the method works for any $d$ but it did not...) by first approximating $1-\varphi(\xi/N)$ with functions of bounded $L^1$-norm and use oscillatory integral. I and my colleagues tried to modify the argument but we failed. Thank you very much for any help!
The professor I mentioned in the question gave a very simple proof. Using Young's inequality, one can easily prove $$ \|P_M |\nabla|^{-s}f \|_{L^p(\mathbb R^d)} \leq C_{p,s,d} M^{-s} \| f\|_{L^p(\mathbb R^d)}$$ where $P_M f = \mathcal F^{-1} ((\varphi (\xi/ M) - \varphi(2\xi/M))\mathcal F(f)(\xi))$. This implies $$\|P_{\geq N}|\nabla|^{-s}f \|_{L^p(\mathbb R^d)}\leq \sum_{M > N:\text{dyadic}} \|P_M|\nabla|^{-s}f \|_{L^p(\mathbb R^d)}\leq C_{p,s,d}\sum_{M > N:\text{dyadic}} M^{-s} \|f \|_{L^p(\mathbb R^d)} \sim N^{-s} \|f \|_{L^p(\mathbb R^d)}.$$ Putting $f = P_{\geq 2N} f$ gives the desired inequality.