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 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 17/May/17
![if: a=b=c⇒RHS=1≮LHS=1 1+((b−c)/a)=2×((p−c)/a),1+((c−a)/b)=2((p−a)/b),1+((a−b)/c)=2((p−b)/c) (RHS)_1 =(1+((b−c)/a))(1+((c−a)/b))(1+((a−b)/c))⇒ (RHS)_1 =8×(((p−a)(p−b)(p−c))/(abc))=8×((S^2 /p)/(4RS))= =((2S)/(pR))=((2r)/R)≤1 (according to Euler′s teorem) d^2 =R^2 −2Rr=R(R−2r)≥0⇒R≥2r abc=((2S)/(sinC)).2RsinC=4R.S, S=p.r (1+((b−c)/a))^a (1+((c−a)/b))^b (1+((a−b)/c))^c < (1+((b−c)/a))^(abc) (1+((c−a)/b))^(abc) (1+((a−b)/c))^(abc) = [(1+((b−c)/a))(1+((c−a)/b))(1+((a−b)/c))]^(abc) = =(((2r)/R))^(abc) ≤1^(abc) ≤1 .■](https://www.tinkutara.com/question/Q13233.png)
$${if}:\:\:{a}={b}={c}\Rightarrow{RHS}=\mathrm{1}\nless{LHS}=\mathrm{1} \\ $$$$\mathrm{1}+\frac{{b}−{c}}{{a}}=\mathrm{2}×\frac{{p}−{c}}{{a}},\mathrm{1}+\frac{{c}−{a}}{{b}}=\mathrm{2}\frac{{p}−{a}}{{b}},\mathrm{1}+\frac{{a}−{b}}{{c}}=\mathrm{2}\frac{{p}−{b}}{{c}} \\ $$$$\left({RHS}\right)_{\mathrm{1}} =\left(\mathrm{1}+\frac{{b}−{c}}{{a}}\right)\left(\mathrm{1}+\frac{{c}−{a}}{{b}}\right)\left(\mathrm{1}+\frac{{a}−{b}}{{c}}\right)\Rightarrow \\ $$$$\left({RHS}\right)_{\mathrm{1}} =\mathrm{8}×\frac{\left({p}−{a}\right)\left({p}−{b}\right)\left({p}−{c}\right)}{{abc}}=\mathrm{8}×\frac{\frac{{S}^{\mathrm{2}} }{{p}}}{\mathrm{4}{RS}}= \\ $$$$=\frac{\mathrm{2}{S}}{{pR}}=\frac{\mathrm{2}{r}}{{R}}\leqslant\mathrm{1}\:\:\left({according}\:{to}\:{Euler}'{s}\:{teorem}\right) \\ $$$${d}^{\mathrm{2}} ={R}^{\mathrm{2}} −\mathrm{2}{Rr}={R}\left({R}−\mathrm{2}{r}\right)\geqslant\mathrm{0}\Rightarrow{R}\geqslant\mathrm{2}{r} \\ $$$${abc}=\frac{\mathrm{2}{S}}{{sinC}}.\mathrm{2}{RsinC}=\mathrm{4}{R}.{S},\:\:{S}={p}.{r} \\ $$$$\left(\mathrm{1}+\frac{{b}−{c}}{{a}}\right)^{{a}} \left(\mathrm{1}+\frac{{c}−{a}}{{b}}\right)^{{b}} \left(\mathrm{1}+\frac{{a}−{b}}{{c}}\right)^{{c}} < \\ $$$$\left(\mathrm{1}+\frac{{b}−{c}}{{a}}\right)^{{abc}} \left(\mathrm{1}+\frac{{c}−{a}}{{b}}\right)^{{abc}} \left(\mathrm{1}+\frac{{a}−{b}}{{c}}\right)^{{abc}} = \\ $$$$\left[\left(\mathrm{1}+\frac{{b}−{c}}{{a}}\right)\left(\mathrm{1}+\frac{{c}−{a}}{{b}}\right)\left(\mathrm{1}+\frac{{a}−{b}}{{c}}\right)\right]^{{abc}} = \\ $$$$=\left(\frac{\mathrm{2}{r}}{{R}}\right)^{{abc}} \leqslant\mathrm{1}^{{abc}} \leqslant\mathrm{1}\:\:\:\:\:.\blacksquare \\ $$