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Question Number 13223 by Tinkutara last updated on 16/May/17

If a, b, c are sides of triangle show that (1 + ((b−c)/a))^a (1 + ((c−a)/b))^b (1 + ((a−b)/c))^c < 1

Ifa,b,caresidesoftriangleshowthat (1+b−ca)a(1+c−ab)b(1+a−bc)c<1

Commented byb.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 .■

if:a=b=c⇒RHS=1≮LHS=1 1+b−ca=2×p−ca,1+c−ab=2p−ab,1+a−bc=2p−bc (RHS)1=(1+b−ca)(1+c−ab)(1+a−bc)⇒ (RHS)1=8×(p−a)(p−b)(p−c)abc=8×S2p4RS= =2SpR=2rR⩽1(accordingtoEuler′steorem) d2=R2−2Rr=R(R−2r)⩾0⇒R⩾2r abc=2SsinC.2RsinC=4R.S,S=p.r (1+b−ca)a(1+c−ab)b(1+a−bc)c< (1+b−ca)abc(1+c−ab)abc(1+a−bc)abc= [(1+b−ca)(1+c−ab)(1+a−bc)]abc= =(2rR)abc⩽1abc⩽1.◼

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