**Proof plan.** Both inequalities reduce to pointwise comparisons between the weights $|k|^{2s}$ and $(1+|k|^2)^s$ on $\mathbb{Z}^n \setminus \{0\}$. The lower bound uses $|k|^2 \leq 1+|k|^2$; the upper bound uses the key torus fact that $|k| \geq 1$ for all $k \neq 0$, which forces $1 \leq |k|^2$ and hence $1 + |k|^2 \leq 2|k|^2$. **Step 1: Lower bound.** For $k \neq 0$ and $s \geq 0$, the inequality $|k|^2 \leq 1+|k|^2$ gives $(|k|^2)^s \leq (1+|k|^2)^s$, i.e., $|k|^{2s} \leq (1+|k|^2)^s$. Since $\hat{f}(0) = 0$, summing over $k \neq 0$: \begin{align*} \|f\|_{\dot{H}^s(\mathbb{T}^n)}^2 = \sum_{k \neq 0} |k|^{2s}|\hat{f}(k)|^2 \leq \sum_{k \neq 0} (1+|k|^2)^s|\hat{f}(k)|^2 = \|f\|_{H^s(\mathbb{T}^n)}^2. \end{align*} **Step 2: Upper bound.** For $k \in \mathbb{Z}^n \setminus \{0\}$, we have $|k|^2 \geq 1$ (since the smallest nonzero value of $|k|^2$ for $k \in \mathbb{Z}^n$ is $1$). Therefore $1 \leq |k|^2$, giving \begin{align*} 1 + |k|^2 \leq |k|^2 + |k|^2 = 2|k|^2. \end{align*} Raising to the power $s \geq 0$: $(1+|k|^2)^s \leq 2^s|k|^{2s}$. Summing over $k \neq 0$ with $\hat{f}(0) = 0$: \begin{align*} \|f\|_{H^s(\mathbb{T}^n)}^2 = \sum_{k \neq 0} (1+|k|^2)^s|\hat{f}(k)|^2 \leq 2^s \sum_{k \neq 0} |k|^{2s}|\hat{f}(k)|^2 = 2^s\|f\|_{\dot{H}^s(\mathbb{T}^n)}^2. \end{align*} Taking square roots gives $\|f\|_{H^s} \leq 2^{s/2}\|f\|_{\dot{H}^s}$. **Step 3: Sharpness.** For $n = 1$ and $s = 1$, consider $f_N(x) = e^{iNx}$ for $N \geq 1$. Then $\hat{f_N}(N) = 1$ and all other Fourier coefficients vanish. The norms are: \begin{align*} \|f_N\|_{\dot{H}^1(\mathbb{T})}^2 = N^2, \qquad \|f_N\|_{H^1(\mathbb{T})}^2 = (1+N^2). \end{align*} The ratio $\|f_N\|_{H^1}^2 / \|f_N\|_{\dot{H}^1}^2 = (1+N^2)/N^2 = 1 + N^{-2} \to 1$ as $N \to \infty$. The supremum of this ratio over all nonzero mean-zero $f$ equals $\sup_{k \neq 0}(1+k^2)/k^2 = (1+1)/1 = 2$ (attained at $k = \pm 1$), confirming that the constant $2^{1/2} = \sqrt{2}$ is sharp.