In The Middle Of 2016!
2 x + 3 x + 4 x + 5 x + 2 0 1 6 + 5 − x + 4 − x + 3 − x + 2 − x
Let x be a real number . Find the minimum value of the expression above.
The answer is 2024.
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2 solutions
Moderator note:
Simple standard approach.
Asides from the preventing a nasty radical, why did excluding 2016 affect the inequality versus including 2016. With 2016 included you should have 9*(2016)^(1/9)
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Constant should not be included. Because constant like 2016 does not depend on x . The maximum and minimum of the rest terms with x should not be dependant on constant. For we can split 2016 into two 1008+1008, then the answer will be 1 0 ( 1 0 0 8 ) 1 / 1 0 , and we split it three parts and different answer again.
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Ah thank you very much sir!
shouldn't it be 1008^2 ?
Wonderful explanation! I was wondering about the same thing.
Thank you so much! Very well explained
And moreover.... if you add 2016 it leads to a negative value on RHS while AM-GM is purely based on the sole root that all a1,a2,....an numbers are positive only I know the answer is a bit late , but ok
Remember that you are solving for the smallest number. Then you might also split the number 2016 into ones (1+1+1... (2016 times)) therefore increasing the denominator to 2024. This is because the equality only holds if and only if all the values are the same. (Also you need to keep in mind that this is an inequality ((2016)^(1/2) is smaller than 2^x+3^x..., but it can never be equal to the term above).
But these numbers are not in A.P, so how do we know whether to apply the am GM inequality or not?
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AP-GM inequality has nothing to do with AP or GP but their means. Read the wiki.
Simply BRILLIANT!!!
Neat and colorful, +1!
here is my solution, it isn't totally rigorous but it's intuitive without loss of generality: assume x ≥ 0 if x > 0 then the terms with positive x's would get way bigger therefore minimizing x would lead to a minimum solution that is x = 0 giving us the answer 2 0 1 6 + 8 = 2 0 2 4
NOTE: if the function isn't increasing on the interval [ 0 , ∞ ) then the proof doesn't hold and that's why it isn't fully rigorous
Relevant wiki: Applying the Arithmetic Mean Geometric Mean Inequality
Applying AM-GM inequality to the following whose all terms are positive real numbers:
2 x + 3 x + 4 x + 5 x + 5 − x + 4 − x + 3 − x + 2 − x ⟹ 2 x + 3 x + 4 x + 5 x + 2 0 1 6 + 5 − x + 4 − x + 3 − x + 2 − x ≥ 8 8 1 = 8 ≥ 2 0 1 6 + 8 = 2 0 2 4