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Question Number 164156 by HongKing last updated on 14/Jan/22

if  ABC  is a triangle with usual  notations, then prove the following  inequality:  (4R + r)^3  - 4r^2 (2R - r) - 3s^2 (2R + r) ≥ 0

$$\mathrm{if}\:\:\mathrm{ABC}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{with}\:\mathrm{usual} \\ $$$$\mathrm{notations},\:\mathrm{then}\:\mathrm{prove}\:\mathrm{the}\:\mathrm{following} \\ $$$$\mathrm{inequality}: \\ $$$$\left(\mathrm{4R}\:+\:\mathrm{r}\right)^{\mathrm{3}} \:-\:\mathrm{4r}^{\mathrm{2}} \left(\mathrm{2R}\:-\:\mathrm{r}\right)\:-\:\mathrm{3s}^{\mathrm{2}} \left(\mathrm{2R}\:+\:\mathrm{r}\right)\:\geqslant\:\mathrm{0} \\ $$

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