2根据余弦定理和中线定理有$$b^2+c^2-2bccos A=2leftfraca^22-fracb^24right+2leftfraca^22-fracc^24right$$$$Rightarrow b^2+c^2-2bccos A=a^2-b^2-c^2$$$$Rightarrow b^2+c^2-2bccos A=2bccos A-2b^2sin^2 C-2c^2sin^2 B$$$$Ri

According to the cosine theorem and the median theorem, we have $$b^2+c^2-2bc\cos A=2\left(\frac{a^2}{2}-\frac{b^2}{4}\right)+2\left(\frac{a^2}{2}-\frac{c^2}{4}\right)$$$$\Rightarrow b^2+c^2-2bc\cos A=a^2-b^2-c^2$$$$\Rightarrow b^2+c^2-2bc\cos A=2bc\cos A-2b^2\sin^2 C-2c^2\sin^2 B$$$$\Rightarrow b^2+2bc\cos A+c^2=2bc\cos A+2b^2\sin^2 C+2c^2\sin^2 B$$$$\Rightarrow (b+c\cos A)^2=b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C+2$$$$\Rightarrow b+\sqrt{2}c\cos A\in[\sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C},\sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C+2}]$$Note that $b\sin C+c\sin B\leq b+c$, so $$b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C\leq (b\sin C+c\sin B)^2\leq (b+c)^2$$$$\Rightarrow \sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C}\leq b+c$$$$\Rightarrow b+\sqrt{2}c\cos A\in[\sqrt{b^2\sin^2 C+c^2\sin^2 B+2bc\sin^2 B\sin^2 C},b+c\sqrt{2}]$$ The above content translated into numerical mode is