证明：
\[ \left\|\mathbf{M} \left(\frac{\bm{s}_1 + \bm{s}_2}{2}\right) - \bm{c}\right\|_2^2 \leq \frac{\|\mathbf{M} \bm{s}_1 - \bm{c}\|_2^2 + \|\mathbf{M} \bm{s}_2 - \bm{c}\|_2^2}{2}. \]
注意到
\[ \begin{split} \left\|\mathbf{M} \left(\frac{\bm{s}_1 + \bm{s}_2}{2}\right) - \bm{c}\right\|_2^2 &= \left(\mathbf{M} \left(\frac{\bm{s}_1 + \bm{s}_2}{2}\right) - \bm{c}\right)^{\mathrm{T}} \left(\mathbf{M} \left(\frac{\bm{s}_1 + \bm{s}_2}{2}\right) - \bm{c}\right) \\ &= \left(\frac{\mathbf{M} \bm{s}_1 + \mathbf{M} \bm{s}_2}{2} - \bm{c}\right)^{\mathrm{T}} \left(\frac{\mathbf{M} \bm{s}_1 + \mathbf{M} \bm{s}_2}{2} - \bm{c}\right) \\ &= \frac{1}{4} \boldsymbol{s}_1 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_1 + \frac{1}{4} \boldsymbol{s}_2 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_2 + \frac{1}{2} \boldsymbol{s}_1 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_2 - (\boldsymbol{s}_1 + \boldsymbol{s}_2)^{\mathrm{T}} \mathbf{M}^{\mathrm{T}} \bm{c} + \bm{c}^{\mathrm{T}} \bm{c} \\ \end{split} \]
而
\[ \begin{split} \|\mathbf{M} \bm{s}_1 - \bm{c}\|_2^2 + \|\mathbf{M} \bm{s}_2 - \bm{c}\|_2^2 &= (\mathbf{M} \bm{s}_1 - \bm{c})^{\mathrm{T}} (\mathbf{M} \bm{s}_1 - \bm{c}) + (\mathbf{M} \bm{s}_2 - \bm{c})^{\mathrm{T}} (\mathbf{M} \bm{s}_2 - \bm{c}) \\ &= \boldsymbol{s}_1 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_1 + \boldsymbol{s}_2 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_2 - 2 \boldsymbol{s}_1^{\mathrm{T}} \mathbf{M}^{\mathrm{T}} \bm{c} - 2 \boldsymbol{s}_2^{\mathrm{T}} \mathbf{M}^{\mathrm{T}} \bm{c} + 2 \bm{c}^{\mathrm{T}} \bm{c}. \end{split} \]
因此只需证明
\[ \begin{split} & \frac{1}{4} \boldsymbol{s}_1 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_1 + \frac{1}{4} \boldsymbol{s}_2 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_2 + \frac{1}{2} \boldsymbol{s}_1 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_2 - (\boldsymbol{s}_1 + \boldsymbol{s}_2)^{\mathrm{T}} \mathbf{M}^{\mathrm{T}} \bm{c} + \bm{c}^{\mathrm{T}} \bm{c} \\ \leq & \frac{1}{2}\boldsymbol{s}_1 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_1 + \frac{1}{2} \boldsymbol{s}_2 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_2 - \boldsymbol{s}_1^{\mathrm{T}} \mathbf{M}^{\mathrm{T}} \bm{c} - \boldsymbol{s}_2^{\mathrm{T}} \mathbf{M}^{\mathrm{T}} \bm{c} + \bm{c}^{\mathrm{T}} \bm{c}. \end{split} \]
消除公共项，只需证明
\[ \begin{split} \frac{1}{4} \boldsymbol{s}_1 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_1 + \frac{1}{4} \boldsymbol{s}_2 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_2 - \frac{1}{2} \boldsymbol{s}_1 \mathbf{M}^{\mathrm{T}} \mathbf{M} \boldsymbol{s}_2 \geq 0. \end{split} \]
上面的不等式可以写成
\[ (\boldsymbol{s}_1 - \boldsymbol{s}_2)^{\mathrm{T}} \mathbf{M}^{\mathrm{T}} \mathbf{M} (\boldsymbol{s}_1 - \boldsymbol{s}_2) \geq 0. \]
也即
\[ \left( \mathbf{M} (\boldsymbol{s}_1 -\boldsymbol{s}_2)\right)^{\mathrm{T}} \left( \mathbf{M} (\boldsymbol{s}_1 -\boldsymbol{s}_2)\right) \geq 0. \]
得证。