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Three numbers (Posted on 2003-11-17) Difficulty: 3 of 5
If x, y and z are real numbers such that: x + y + z = 5 and xy + yz + zx = 3, what is the largest value that x can have ?

No Solution Yet Submitted by Ravi Raja    
Rating: 3.7500 (4 votes)

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Solution Solution by Cauchy-Schwarz | Comment 19 of 21 |

(x + y + z) = x + y + z + 2(xy + yz + zx)
Therefore x + y + z = 19

y + z = 5 - x
y + z = 19 - x

By the Cauchy-Schwarz inequality,
|y + z| <= (1 + 1) (y + z), with equality if and only if y = z.

Hence |5 - x| <= (2(19 - x)).
Squaring, we obtain x - 10x + 25 <= 38 - 2x, and so
3x - 10x - 13 = (x + 1)(3x - 13) <= 0.

One factor must be positive and the other negative; hence -1 <= x <= 13/3.

Verifying that the maximum and minimum may be attained: x = 13/3, y = z = 1/3; x = -1, y = z = 3.


  Posted by Nick Hobson on 2004-11-12 15:56:09
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