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Question Number 13236 by Abbas-Nahi last updated on 17/May/17

$$Forpositivea,b,csuchthatabc=1$$$$showthat{a}^{b+c}{b}^{c+a}{c}^{a+b}⩽1$$$$solution:$$$$a^{b+c}{b}^{c+a}{c}^{a+b}=\left(a^b{a}^c{b}^c{b}^a{c}^a{c}^b\right)$$$$={\left(b\times c\right)}^a{\left(a\times c\right)}^b{\left(a\times b\right)}^c$$$$={\left(a^0\times b\times c\right)}^a{\left(a\times b^0\times c\right)}^b{\left(a\times b\times c^0\right)}^c$$$$⩽{\left(a\times b\times c\right)}^a{\left(a\times b^{}\times c\right)}^b{\left(a\times b\times c\right)}^c$$$$⩽{\left(1\right)}^a{\left(1\right)}^b{\left(1\right)}^c;sinceabc=1$$$$⩽1$$

Commented by RasheedSindhi last updated on 17/May/17

$$={\left(a^0\times b\times c\right)}^a{\left(a\times b^0\times c\right)}^b{\left(a\times b\times c^0\right)}^c$$$$⩽{\left(a\times b\times c\right)}^a{\left(a\times b^{}\times c\right)}^b{\left(a\times b\times c\right)}^c$$$$?$$$$\mathrm{Do}{a}^a{b}^b{c}^c⩾1?$$$$$$

Commented by mrW1 last updated on 17/May/17

$$howdidyouget$$$${\left(a^0\times b\times c\right)}^a$$$$⩽{\left(a\times b\times c\right)}^a$$$$?$$$$sinceifa<1,wehave{a}^a<1and$$$${\left(a^0\times b\times c\right)}^a>{\left(a\times b\times c\right)}^a$$

Commented by Tinkutara last updated on 17/May/17

$$\mathrm{Yes},\mathrm{your}\mathrm{solution}\mathrm{is}\mathrm{incorrect}.\mathrm{Please}$$$$\mathrm{check}\mathrm{it}.\mathrm{I}\mathrm{now}\mathrm{found}\mathrm{a}\mathrm{better}\mathrm{solution}.$$$$\mathrm{From}\mathrm{the}\mathrm{given}\mathrm{expression}$$$$a^{b+c}{b}^{c+a}{c}^{a+b}=a^{b+c}{\left(\frac{1}{ac}\right)}^{c+a}{c}^{a+b}[\because abc=1]$$$$=a^{b+c}{a}^{-(c+a)}{c}^{-(c+a)}{c}^{a+b}$$$$=\frac{a^{b+c-c-a}}{c^{c+a-a-b}}=\frac{a^{b-a}}{c^{c-b}}.$$$$\mathrm{We}\mathrm{can}\mathrm{assume}\mathrm{that}a<b<c\mathrm{as}\mathrm{the}$$$$\mathrm{equation}\mathrm{is}\mathrm{symmetric}.$$$$\mathrm{So}\mathrm{the}\mathrm{numbers}\mathrm{are}\mathrm{positive}\mathrm{and}\mathrm{raised}$$$$\mathrm{to}\mathrm{positive}\mathrm{powers}.$$$$\mathrm{So}\mathrm{the}\mathrm{denominator}\mathrm{is}\mathrm{greater}\mathrm{than}\mathrm{the}$$$$\mathrm{numerator}.$$$$\mathrm{Hence}\mathrm{the}\mathrm{above}\mathrm{expression}\mathrm{is}⩽1.$$

Commented by RasheedSindhi last updated on 17/May/17

$$\mathrm{Mr}.\mathrm{Tinkutara}$$$$‘‘\mathrm{So}\mathrm{the}\mathrm{numbers}\mathrm{are}\mathrm{positive}{\underset{-}{\mathrm{integers}}}^{″}$$$$\mathrm{The}\mathrm{given}\mathrm{codition}\mathrm{is}\mathrm{only}$$$$a,b,c>0\wedge abc=1$$$$\mathrm{Why}{\mathrm{you}}^{\prime }\mathrm{ve}\mathrm{assumed}a,b,c\mathrm{as}$$$$\mathbf{integers}?$$

Commented by Abbas-Nahi last updated on 17/May/17

$$sorrysirsince{a}^0⩽a^{1}then$$$${\left(a^0\times b\times c\right)}^a⩽{\left(a^1\times b\times c\right)}^a$$$$\left(bycomparing\right)$$$$$$

Commented by Abbas-Nahi last updated on 17/May/17

$$then{\left(a^0\times b\times c\right)}^a⩽\left(1\right)$$$$since{\left(a^1\times b\times c\right)}^a={\left(1\right)}^a=1$$$$$$

Commented by Abbas-Nahi last updated on 17/May/17

$$ToMr.Rasheed$$$$mysirsince{a}^0⩽a^1,b^0⩽b^1,c^0⩽c^{1}then$$$${\left(a^0\times b\times c\right)}^a{\left(a\times b^0\times c\right)}^b{\left(a\times b\times c^0\right)}^c$$$$⩽{\left(a\times b\times c\right)}^a{\left(a\times b\times c\right)}^b{\left(a\times b\times c\right)}^c$$$$\left(comparing\right)andsince(a\times b\times c=1)$$$$⩽{\left(1\right)}^a{\left(1\right)}^b{\left(1\right)}^c$$$$⩽\left(1\right)\left(1\right)\left(1\right)$$$$⩽\left(1\right)$$$$finished$$$$$$

Commented by RasheedSindhi last updated on 19/May/17

$$\mathrm{But}\mathrm{not}\mathrm{necessarily}$$$$a^0⩽a^1$$$$\mathrm{For}\mathrm{example}\mathrm{if}a=1/2$$$${\left(1/2\right)}^0\stackrel{?}{⩽}{\left(1/2\right)}^1$$$$1⪇\frac{1}{2}$$$$\mathrm{Actually}\mathrm{for}\mathrm{any}\mathrm{positive}a$$$$\mathrm{less}\mathrm{than}1:$$$$a^0⪇a^1$$

Commented by Abbas-Nahi last updated on 17/May/17

$$sorrysir;thequestionandtherelation$$$$aboveareTrueifandonlyifa=0\wedge a=1$$

Commented by mrW1 last updated on 17/May/17

$$ifa=0\Rightarrow abc=0,butabc=1!$$

Commented by Tinkutara last updated on 17/May/17

$$\mathrm{We}\mathrm{cannot}\mathrm{take}a=0\mathrm{because}\mathrm{it}\mathrm{will}$$$$\mathrm{violate}\mathrm{the}\mathrm{condition}abc=1.$$

Commented by RasheedSindhi last updated on 17/May/17

$$a=0\wedge a=1\mathrm{means}a\mathrm{is}\mathrm{simultaneously}$$$$\mathrm{equal}\mathrm{to}0\mathrm{\&}1?$$$$\mathrm{Given}\mathrm{condition}:abc=1$$$$a,b,c\mathrm{may}\mathrm{take}\mathrm{any}\mathrm{real}\mathrm{value}$$$$\mathrm{other}\mathrm{than}0.$$$$\mathrm{For}\mathrm{example}:$$$$\left(1/2\right)\left(6\right)\left(1/3\right)$$$$\mathrm{To}\mathrm{Mr}.\mathrm{Abbas}$$$$\mathrm{pl}\mathrm{set}a=1/2,b=6\mathrm{and}c=1/3$$$$\mathrm{in}\mathrm{your}\mathrm{proof}\mathrm{and}\mathrm{see}!$$$$(\mathrm{These}\mathrm{values}\mathrm{fulfil}abc=1)$$

Commented by Tinkutara last updated on 17/May/17

$$\mathrm{Sorry},\mathrm{it}\mathrm{will}\mathrm{not}\mathrm{be}\mathrm{integers}.$$

Commented by prakash jain last updated on 18/May/17

$$\mathrm{To}\mathrm{Tinkutara}$$$$\mathrm{Your}\mathrm{argument}\mathrm{that}\mathrm{positive}$$$$\mathrm{numbers}\mathrm{raised}\mathrm{to}+\mathrm{ve}\mathrm{power}$$$$\mathrm{hence}\mathrm{denominator}\mathrm{is}\mathrm{greater}$$$$\mathrm{than}\mathrm{numerator}\mathrm{is}\mathrm{not}\mathrm{valid}.$$$$\mathrm{In}\mathrm{general}$$$$a>b,x>0,y>0$$$$\mathrm{does}\mathrm{not}\mathrm{imply}$$$$a^x>b^y$$

Commented by Tinkutara last updated on 19/May/17

$$\mathrm{But}\mathrm{here}\mathrm{it}\mathrm{is}\mathrm{given}\mathrm{that}abc=1.$$

Commented by prakash jain last updated on 19/May/17

$$\mathrm{I}\mathrm{added}\mathrm{a}\mathrm{clarification}$$$$\mathrm{that}\mathrm{the}\mathrm{readers}\mathrm{know}\mathrm{that}\mathrm{an}$$$$\mathrm{additional}\mathrm{condition}\mathrm{is}\mathrm{required}$$$$a<c\mathrm{is}\mathrm{not}\mathrm{sufficient}.$$$$\mathrm{MrW}1\mathrm{added}\mathrm{in}\mathrm{another}\mathrm{post}$$$$\mathrm{that}a⩽1\mathrm{and}c⩾1\mathrm{which}\mathrm{is}\mathrm{an}$$$$\mathrm{additional}\mathrm{condition}.$$

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