Inequality within an inequality
Given that 0 ≤ a ≤ b ≤ c ≤ d ≤ e and a + b + c + d + e = 1 .
Find the maximum value of a b + c d + b c + b e + e a .
This is part of the set My Problems and THRILLER
The answer is 0.2.
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1 solution
Nice solution buddy (+1)... Did the same way.
@Rahil Sehgal Thanks! :) :)
I think you give the inequality direction wrong it should be ≥
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In which step?
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Where you use Chebyshev's inequality
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@Kushal Bose – No its fine . Infact when we take product of oppositely paired numbers , it is at the lowest rung.
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@Ankit Kumar Jain – but in d wiki it is given in opposite direction
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@Kushal Bose – See in the link that I have provided , it is given in the same way.
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@Ankit Kumar Jain – Yes I checked this link and find this.Can u recheck
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@Kushal Bose – I think you are getting confused by this.
In the link it is this a 1 ≥ a 2 ≥ a 3 ⋯ and b 1 ≥ b 2 ≥ b 3 ⋯ and then the conditions are given . But notice that the second set in my solution is written in reverse order and hence the result.
I'm pretty sure there's a typo in the problem. The 1st term in the expression you're maximizing is ab, but in your solution it is ad, which makes a lot more sense.
Consider this :
a ≤ b ≤ c ≤ d ≤ e
d + e ≥ c + e ≥ b + d ≥ a + c ≥ a + b
By Chebyshev's Inequality ;
5 a ( d + e ) + b ( c + e ) + c ( b + d ) + d ( a + c ) + e ( a + b ) ≤ ( 5 a + b + c + d + e ) ( 5 ( d + e ) + ( c + e ) + ( b + d ) + ( a + c ) + ( a + b ) )
⇒ a ( d + e ) + b ( c + e ) + c ( b + d ) + d ( a + c ) + e ( a + b ) ≤ 5 2
2 ( a d + b c + d c + b e + e a ) ≤ 5 2
a d + d c + b c + b e + e a ≤ 5 1
Equality holds when a = b = c = d = e = 5 1