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Question

If logeabc=logebca=logecab, then prove that ab+c+bc+a+ca+b3.

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Solution

Since a>0,b>0 & c>0
Given logeabc=logebca=logeccb

(b+c)logeab2c2=(c+a)logebc2a2=(a+b)logeca2b2=k(say)

(b+c)logea=k(b2c2) ....(1)

(c+a)logeb=k(c2a2) ....(2)

(a+b)logec=k(a2b2) ......(3)

Adding these equations, we get

(b+c)logea+(c+a)logeb+(a+b)logec=k(b2c2+c2a2+a2b2)

(b+c)logea+(c+a)logeb+(a+b)logec=0
logeab+c+logebc+a+logeca+b=0,[xloga=logax]

loge(ab+c.bc+a.ca+b)=0,[loga+logb+logc=log(abc)]

ab+c.bc+a.ca+b=1 ........(4)

AgainA.M.G.M.

ab+c+bc+a+ca+b3(ab+cbc+aca+b)1/3=(1)1/3=1, using (4)

ab+c+bc+a+ca+b3. [Hence proved]


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