What are A,M and C?

Algebra Level 3

Let A , B , C A,B,C be non negative integers such that A + M + C = 12 A+M+C=12 . What is the maximum value of A M C + A M + M C + C A AMC+AM+MC+CA

112 92 72 62

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Vinayak Bansal by Vinayak Bansal , 17, India

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2 solutions

Donglin Loo
May 24, 2018

A M C + A M + M C + C A + A + M + C + 1 = ( A + 1 ) ( M + 1 ) ( C + 1 ) AMC+AM+MC+CA+A+M+C+1=(A+1)(M+1)(C+1) A M C + A M + M C + C A = ( A + 1 ) ( M + 1 ) ( C + 1 ) − 13 AMC+AM+MC+CA=(A+1)(M+1)(C+1)-13

By AM-GM Inequality, ( A + 1 ) ( M + 1 ) ( C + 1 ) ≤ ( A + 1 + M + 1 + C + 1 3 ) 3 = 125 (A+1)(M+1)(C+1)\leq(\cfrac{A+1+M+1+C+1}{3})^3=125

A M C + A M + M C + C A ≤ 125 − 13 = 112 AMC+AM+MC+CA\leq125-13=112

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X X
May 21, 2018

Just put in A = M = C = 4 A=M=C=4

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What is the proof that A M C + A M + M C + C A AMC+AM+MC+CA will be maximum when A = M = C = 4 A=M=C=4 ?

Vinayak Bansal - 3 years ago

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Apply to AM-GM inequality,the maximum value of A M C AMC occurs when A = M = C = 4 A=M=C=4 .

And about A M + M C + C A AM+MC+CA ,because A 2 + M 2 + C 2 ≥ A M + M C + C A , 144 = ( A + M + C ) 2 ≥ 3 ( A M + M C + C A ) A^2+M^2+C^2 \ge AM+MC+CA,144=(A+M+C)^2 \ge 3(AM+MC+CA) ,the maximum value also occurs when A = M = C = 4 A=M=C=4

X X - 3 years ago

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Thank you for the proof !

Vinayak Bansal - 3 years ago

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