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What is the minimum value that the expression
x 2 y + x x 2 y 2 − 2 x 2 y + 2 x 2 + 2 x y − 2 x + 1
when x , y ∈ R + , can attain? Express your answer to four decimal places.
The answer is 0.8284.
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1 solution
Exact same way, it's pretty easy once you factor out everything.
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All those who could and could'nt solve it please try this too , it was asked in JOMO 10 (now as the competition is over so i can discuss it here) A nice question
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is it 2 4 @megh choksi ??
post it as problem please
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@Omar El Mokhtar – I don't have the permission to do as it was asked in JOMO 10 , only the problems creators can do it according to the rules and regulations of JOMO
Suppose we are asked for the smallest value 2 + 8 can take. Surely it would be fallacious to reason that, since the arithmetic mean is greater than or equal to the geometric mean, 2 2 + 8 ≥ 2 ⋅ 8 = 4 , and so 2 + 8 ≥ 8 , and hence the minimum value of 2 + 8 is 8 .
Am I missing something, or do you really need to add to this a demonstration that the value 2 is in fact obtainable (as it indeed is, for any point on the curve y = 2 − x 1 ).
x ( x y + 1 ) x 2 y 2 + 2 x y + 1 − 2 x 2 y + 2 x 2 − 2 x
= x ( x y + 1 ) ( x y + 1 ) 2 − 2 x ( x y + 1 − x )
= x x y + 1 + x y + 1 2 x − 2
A . M ≥ G . M
2 x x y + 1 + x y + 1 2 x ≥ 2
m i n i m u m v a l u e = 2 2
2 2 − 2 = 0 . 8 2 8