Question Number 13223 by Tinkutara last updated on 16/May/17 | ||
$$\mathrm{If}\:{a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{show}\:\mathrm{that} \\ $$ $$\left(\mathrm{1}\:+\:\frac{{b}−{c}}{{a}}\right)^{{a}} \left(\mathrm{1}\:+\:\frac{{c}−{a}}{{b}}\right)^{{b}} \left(\mathrm{1}\:+\:\frac{{a}−{b}}{{c}}\right)^{{c}} \:<\:\mathrm{1} \\ $$ | ||
Commented byb.e.h.i.8.3.4.1.7@gmail.com last updated on 17/May/17 | ||
$${if}:\:\:{a}={b}={c}\Rightarrow{RHS}=\mathrm{1}\nless{LHS}=\mathrm{1} \\ $$ $$\mathrm{1}+\frac{{b}−{c}}{{a}}=\mathrm{2}×\frac{{p}−{c}}{{a}},\mathrm{1}+\frac{{c}−{a}}{{b}}=\mathrm{2}\frac{{p}−{a}}{{b}},\mathrm{1}+\frac{{a}−{b}}{{c}}=\mathrm{2}\frac{{p}−{b}}{{c}} \\ $$ $$\left({RHS}\right)_{\mathrm{1}} =\left(\mathrm{1}+\frac{{b}−{c}}{{a}}\right)\left(\mathrm{1}+\frac{{c}−{a}}{{b}}\right)\left(\mathrm{1}+\frac{{a}−{b}}{{c}}\right)\Rightarrow \\ $$ $$\left({RHS}\right)_{\mathrm{1}} =\mathrm{8}×\frac{\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)}{{abc}}=\mathrm{8}×\frac{\frac{{S}^{\mathrm{2}} }{{p}}}{\mathrm{4}{RS}}= \\ $$ $$=\frac{\mathrm{2}{S}}{{pR}}=\frac{\mathrm{2}{r}}{{R}}\leqslant\mathrm{1}\:\:\left({according}\:{to}\:{Euler}'{s}\:{teorem}\right) \\ $$ $${d}^{\mathrm{2}} ={R}^{\mathrm{2}} −\mathrm{2}{Rr}={R}\left({R}−\mathrm{2}{r}\right)\geqslant\mathrm{0}\Rightarrow{R}\geqslant\mathrm{2}{r} \\ $$ $${abc}=\frac{\mathrm{2}{S}}{{sinC}}.\mathrm{2}{RsinC}=\mathrm{4}{R}.{S},\:\:{S}={p}.{r} \\ $$ $$\left(\mathrm{1}+\frac{{b}−{c}}{{a}}\right)^{{a}} \left(\mathrm{1}+\frac{{c}−{a}}{{b}}\right)^{{b}} \left(\mathrm{1}+\frac{{a}−{b}}{{c}}\right)^{{c}} < \\ $$ $$\left(\mathrm{1}+\frac{{b}−{c}}{{a}}\right)^{{abc}} \left(\mathrm{1}+\frac{{c}−{a}}{{b}}\right)^{{abc}} \left(\mathrm{1}+\frac{{a}−{b}}{{c}}\right)^{{abc}} = \\ $$ $$\left[\left(\mathrm{1}+\frac{{b}−{c}}{{a}}\right)\left(\mathrm{1}+\frac{{c}−{a}}{{b}}\right)\left(\mathrm{1}+\frac{{a}−{b}}{{c}}\right)\right]^{{abc}} = \\ $$ $$=\left(\frac{\mathrm{2}{r}}{{R}}\right)^{{abc}} \leqslant\mathrm{1}^{{abc}} \leqslant\mathrm{1}\:\:\:\:\:.\blacksquare \\ $$ | ||