Question Number 16845 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 27/Jun/17
Commented by prakash jain last updated on 28/Jun/17
$$\left.\mathrm{1}\right) \\ $$$$\mathrm{cos}\:{A}+\mathrm{cos}\:{B}\mathrm{cos}\:{C} \\ $$$${C}=\mathrm{90}°,{A}=\mathrm{45}°,{B}=\mathrm{45}° \\ $$$$\mathrm{cos}\:{A}+\mathrm{cos}\:{B}\centerdot\mathrm{cos}\:{C}=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{cos}\:{A}+\mathrm{cos}\:{B}\centerdot\mathrm{cos}\:{C}}\ngeqslant\mathrm{4} \\ $$
Commented by b.e.h.i.8.3.417@gmail.com last updated on 07/Jul/17
$${in}\:{your}\:{case}: \\ $$$${cosA}+{cosB}.{cosC}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${cosB}+{cosA}.{cosC}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${cosC}+{cosB}.{cosA}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Sigma\frac{\mathrm{1}}{{cosA}+{cosB}.{cosC}}=\mathrm{2}×\sqrt{\mathrm{2}}+\mathrm{2}= \\ $$$$=\mathrm{2}×\mathrm{1}.\mathrm{41}+\mathrm{2}=\mathrm{2}.\mathrm{82}+\mathrm{2}=\mathrm{4}.\mathrm{82}\geqslant\mathrm{4} \\ $$
Commented by prakash jain last updated on 08/Jul/17
$$\mathrm{ok}.\:\mathrm{i}\:\mathrm{understand}\:\mathrm{what}\:\mathrm{u}\:\mathrm{meant}. \\ $$$$\mathrm{i}\:\mathrm{took}\:\mathrm{only}\:\mathrm{one}\:\mathrm{term}. \\ $$
Answered by b.e.h.i.8.3.417@gmail.com last updated on 08/Jul/17
$$\left.\mathrm{1}\right){cosA}+{cosB}.{cosC}={cos}\left(\mathrm{180}−{B}−{C}\right)+{cosB}.{cosC}= \\ $$$$=−{cos}\left({B}+{C}\right)+{cosB}.{cosC}={sinB}.{sinC} \\ $$$$\Rightarrow{LHS}=\Sigma\frac{\mathrm{1}}{{sinA}.{sinB}}=\frac{\Sigma{sinA}}{\Pi{sinA}}= \\ $$$$=\frac{\Sigma\frac{{a}}{\mathrm{2}{R}}}{\Pi\frac{{a}}{\mathrm{2}{R}}}=\frac{\frac{\mathrm{2}{p}}{\mathrm{2}{R}}}{\frac{{abc}}{\mathrm{8}{R}^{\mathrm{3}} }}=\frac{\frac{{p}}{{R}}}{\frac{\mathrm{4}{RS}}{\mathrm{8}{R}^{\mathrm{3}} }}=\frac{{p}}{{S}}.\mathrm{2}{R}=\frac{\mathrm{2}{R}}{{r}}\geqslant\mathrm{2}×\mathrm{2}=\mathrm{4}. \\ $$