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 Three numbers (Posted on 2003-11-17)
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 by Cauchy-Schwarz | Comment 19 of 21 |

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
Therefore x² + y² + z² = 19

y + z = 5 - x
y² + z² = 19 - x²

By the Cauchy-Schwarz inequality,
|y + z| <= (1² + 1²)½ (y² + z²)½, with equality if and only if y = z.

Hence |5 - x| <= (2(19 - x²))½.
Squaring, we obtain x² - 10x + 25 <= 38 - 2x², and so
3x² - 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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