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If x , y , z are real numbers such that x + y + z = 5 and y z + z x + x y = 8 .
Find the minimum value of x .
The answer is 1.
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2 solutions
Yes, that looks good... you do need to verify that x = 1 is attained.
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it can be shown by factoring quadratics
By symmetry, we have y = z when x is minimal (or maximal). The equations x + 2 y = 5 and 2 x y + y 2 = 8 have the two solutions x = 1 , y = 2 and x = 3 7 , y = 3 4 , so that the minimal value of x is 1 .
sir, have a look at my solution
@Otto Bretscher Sir can you please explain me as to how can you say that when x is minimal or maximal , then y = z ???!!!
We can calculate :
x 2 + y 2 + z 2 = ( x + y + z ) 2 − 2 ( x y + y z + z x ) = 9
Then
y + z = 5 − x
y 2 + z 2 = 9 − x 2
Using Cauchy S. inequality,
2 ( y 2 + z 2 ) > ( y + z ) 2
2 ( 9 − x 2 ) > ( 5 − x ) 2
0 > 3 x 2 − 1 0 x + 7
so min is 1 and max is 2.333....