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Algebra Level 4

a 2 b 3 c 4 \large a^2b^3c^4

Positive real numbers a a , b b and c c are such that a + b + c = 18 a+b+c=18 . If the maximum value of expression above can be written as p 1 α p 2 β p_1^\alpha \cdot p_2^\beta , where p 1 p_1 and p 2 p_2 are prime numbers and α \alpha , β \beta are positive integers, find p 1 + p 2 + α + β p_1+p_2+\alpha+\beta .


The answer is 27.

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1 solution

Sabhrant Sachan
Oct 1, 2017

Relevant wiki: Arithmetic Mean - Geometric Mean

By A.M-G.M

( a 2 + a 2 ) + ( b 3 + b 3 + b 3 ) + ( c 4 + c 4 + c 4 + c 4 ) 9 ( a 2 b 3 c 4 2 2 3 3 4 4 ) 1 9 2 9 ( a 2 b 3 c 4 2 2 3 3 4 4 ) a 2 b 3 c 4 3 3 2 19 \dfrac{ \left( \frac{a}{2}+\frac{a}{2} \right) + \left( \frac{b}{3}+\frac{b}{3}+\frac{b}{3} \right) +\left( \frac{c}{4}+\frac{c}{4}+\frac{c}{4}+\frac{c}{4} \right) }{9} \ge \left( \dfrac{a^2 b^3 c^4}{2^2 3^3 4^4 }\right)^{\frac{1}{9}} \\ 2^9 \ge \left( \dfrac{a^2 b^3 c^4}{2^2 3^3 4^4 }\right) \\ a^2 b^3 c^4 \le 3^3 2^{19 }

and for this maximum to exist a 2 = b 3 = c 4 \dfrac{a}{2} = \dfrac{b}{3} = \dfrac{c}{4}

a = 4 a=4 , b = 6 b=6 , c = 8 c=8

Its simply weighted AM-GM

Md Zuhair - 3 years, 8 months ago

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