The Answer Is Not 0.225
What is the smallest real number k (to 3 decimal places), such that for all ordered triples of non-negative reals ( a , b , c ) which satisfy a + b + c = 1 , we have
1 − c a + 1 − a b + 1 − b c ≤ 1 + k ?
The answer is 0.250.
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1 solution
Could you please explain how maximum is achieved at (0,3/4,1/4)?
did you use partial differentiation?
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Currently, the only method that I know of is to use lagrange multipliers. This is not an easy problem to approach, and the unequal roots makes it harder to use classical methods.
how did you got (0, 3/4, 1/4) ?
I do have a nice algebraic proof but its somewhat long .
Why not .225 (with 1/3,1/3,1/3)
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we are finding max value for 'k'.. thats y. not 0.225 but 0.250.
with (1/2,1/2,0) we get 1.207 can you explain the solution? i went on with random values for a,b and c. which gives k=0.207
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As remarked, substituting in (0, 3/4, 1/4) gives us the value of 0.25, and hence k ≥ 0 . 2 5 .
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oh yes.. my bad.. we have to find maximum.. :D sorry! but can u give out the steps to get the answer? please.?
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@Nisarg Thakkar – Calvin replied above that his key was to use Lagrange multipliers... yeah it's going to be a while till someone figures out a method from that.
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@Jared Low – yea.. when i was in my school, my teacher mentioned it a couple of times. But i dont remember it now.. true, we have to wait..
[This is not a solution.]
The maximum is achieved at ( 0 , 4 3 , 4 1 ) and cyclic permutations, which gives k = 0 . 2 5 0 .
Now it's up to you to prove that
1 − c a + 1 − a b + 1 − b c ≤ 1 . 2 5 .