An algebra problem by Saurabh Patil

Algebra Level 3

Let a , b , c a,b,c be positive numbers that sum to 5. The greatest possible value of a 3 b 2 c 4 a^3 \cdot b^2 \cdot c^4 can be expressed as 2 x 3 y 5 z , 2^x \cdot 3^y \cdot 5^z, where x , y , z x,y,z are integers. Find x + y + z . x+y+z.


The answer is 4.

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1 solution

Andy Hayes
Jul 19, 2017

The strategy is to use the weighted AM-GM inequality . Assign coefficients to the variables that correspond to the exponents:

a 1 3 b 1 2 c 1 4 \begin{array}{l} a \Rightarrow \frac{1}{3} \\ b \Rightarrow \frac{1}{2} \\ c \Rightarrow \frac{1}{4} \end{array}

Now weighted AM-GM inequality:

3 ( 1 3 a ) + 2 ( 1 2 b ) + 4 ( 1 4 c ) 3 + 2 + 4 ( 1 3 a ) 3 ( 1 2 b ) 2 ( 1 4 c ) 4 3 + 2 + 4 a + b + c 3 2 a 3 b 2 c 4 2 10 3 3 9 5 9 3 18 a 3 b 2 c 4 2 10 3 3 2 10 3 15 5 9 a 3 b 2 c 4 \begin{aligned} \frac{3\left(\frac{1}{3}a\right)+2\left(\frac{1}{2}b\right)+4\left(\frac{1}{4}c\right)}{3+2+4} &\ge \sqrt[3+2+4]{\left(\frac{1}{3}a\right)^3 \left(\frac{1}{2}b\right)^2 \left(\frac{1}{4}c\right)^4} \\ \\ \frac{a+b+c}{3^2} &\ge \sqrt[9]{\frac{a^3b^2c^4}{2^{10} \cdot 3^3}} \\ \\ \frac{5^9}{3^{18}} &\ge \frac{a^3b^2c^4}{2^{10} \cdot 3^3} \\ \\ 2^{10} \cdot 3^{-15} \cdot 5^9 &\ge a^3b^2c^4 \end{aligned}

Thus, the maximum is 2 10 3 15 5 9 , 2^{10} \cdot 3^{-15} \cdot 5^9, and x + y + z = 4 . x+y+z=\boxed{4}.

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