An algebra problem by Dinesh Chavan
Let x , y , z be positive real numbers.
Given that x y z = 1
Then find the minimum value of
x 2 + 3 x + 2 1 + y 2 + 3 y + 2 1 + z 2 + 3 z + 2 1
The answer is 0.5.
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2 solutions
@T B I converted your comment into a solution. Can you provide more details on how to proceed?
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Using AM-GM inequality, we gather that 3 x + y + z ≥ 3 x y z = 3 1 = 1 Since we want the minimum value, 3 x + y + z = 1 which can only be true if and only if x = y = z Thus, x = y = z = 1 Fitting this values in, we get, 1 + 3 + 2 1 + 1 + 3 + 2 1 + 1 + 3 + 2 1 = 6 3 = 2 1 = 0 . 5
How does x + y + z having the least value possible make c y c ∑ x 2 + 3 x + 2 1 reach its minimum value?
That's not a proof at all...
The function remains invariant under permutation of the variables, and so we may introduce an ordering. Then we can proceed using Chebyshev's mean inequality, use AM-HM on the factors and finally AM-GM as correctly stated above.
Edit: of course, the equality case is generally easy to guess, so as we can see the problem can be solved without actually solving it.
The function remains invariant under permutation of the variables, and so we may introduce an ordering. Then we can proceed using Chebyshev's mean inequality, use AM-HM on the factors and finally AM-GM as correctly stated above. Edit: of course, the equality case is generally easy to guess, so as we can see the problem can be solved without actually solving it.