三角形边长与角的余弦值之间的关系证明

根据余弦定理和中线定理，有/n$$b^2+c^2-2bc/cos A=2/left(/frac{a^2}{2}-/frac{b^2}{4}/right)+2/left(/frac{a^2}{2}-/frac{c^2}{4}/right)$$/n$$/n$$/Rightarrow b^2+c^2-2bc/cos A=a^2-b^2-c^2$$/n$$/n$$/Rightarrow b^2+c^2-2bc/cos A=2bc/cos A-2b^2/sin^2 C-2c^2/sin^2 B$$/n$$/n$$/Rightarrow b^2+2bc/cos A+c^2=2bc/cos A+2b^2/sin^2 C+2c^2/sin^2 B$$/n$$/n$$/Rightarrow (b+c/cos A)^2=b^2/sin^2 C+c^2/sin^2 B+2bc/sin^2 B/sin^2 C+2$$/n$$/n$$/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}]$$/n/n注意到$b/sin C+c/sin B/leq b+c$，因此/n$$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$$/n$$/n$$/Rightarrow /sqrt{b^2/sin^2 C+c^2/sin^2 B+2bc/sin^2 B/sin^2 C}/leq b+c$$/n$$/n$$/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}]$$