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Question Number 77346 by BK last updated on 05/Jan/20

Commented by BK last updated on 05/Jan/20

$$\mathrm{prove}\mathrm{that}$$

Commented by mind is power last updated on 05/Jan/20

$$\mathrm{squart}\Rightarrow {\mathrm{a}}^2+\mathrm{ab}+\mathrm{bc}+{\mathrm{c}}^2+{2\mathrm{b}}^2+2\sqrt{{\mathrm{a}}^2+\mathrm{ab}+{\mathrm{b}}^2}.\sqrt{{\mathrm{b}}^2+\mathrm{bc}+{\mathrm{c}}^2}>{\mathrm{a}}^2+{\mathrm{c}}^2+\mathrm{ac}$$$$\mathrm{evident}\mathrm{since}\sqrt{{\mathrm{b}}^2+\mathrm{bc}+{\mathrm{c}}^2}⩾\mathrm{c}\mathrm{and}\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+\mathrm{ab}}⩾\mathrm{a}$$

Commented by BK last updated on 05/Jan/20

$$\mathrm{thank}\mathrm{you}$$

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