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Question Number 183167 by Shrinava last updated on 21/Dec/22

$$\mathrm{In}△\mathrm{ABC}\mathrm{the}\mathrm{following}\mathrm{relationship}$$$$\mathrm{holds}:$$$$\frac{{\mathrm{m}}_{\mathbf{b}}}{\mathrm{b}}+\frac{{\mathrm{m}}_{\mathbf{c}}}{\mathrm{c}}⩽\frac{\mathrm{a}}{2\mathrm{r}}⩽\frac{{\mathrm{n}}_{\mathbf{b}}}{{\mathrm{h}}_{\mathbf{b}}}+\frac{{\mathrm{n}}_{\mathbf{c}}}{{\mathrm{h}}_{\mathbf{c}}}$$

Commented by mr W last updated on 22/Dec/22

$$ithinknoteverybodyknowswhatis$$$$whatintheequation.$$

Commented by Frix last updated on 22/Dec/22

$${\mathrm{That}}^{\prime }\mathrm{s}\mathrm{plain}\mathrm{to}\mathrm{see}\mathrm{because}\mathrm{obviously}$$$$(\frac{\alpha }{{\rho }_b+{\rho }_c}+\frac{\beta }{{\rho }_a+{\rho }_c})(i_c-j_c)>\frac{\gamma }{{\rho }_a+{\rho }_b}((i_a-j_a)+(i_b-j_b))$$

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