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ღ kekio ღ

21/08/2017 at 20:41
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Proof: a) x ^ 2 + x + 1 > 0 for all x; 
b) 2 x ^ 2 + 2 x + 1 > = 0 for all x



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    Dao Trong Luan 21/08/2017 at 21:18

    a,

    x2 + x + 1

    = x2 + 2.x.1/2 + [1/2]2 + 3/4

    = [x + 1/2]2 + 3/4 \(\ge\) 3/4 

    => x2 + x + 1 > 0

    b,

    2x2 + 2x + 1

    = x2 + x2 + 2.x.1 + 12

    = x2 + [x+1]2 \(\ge\)0 + 0 = 0

    => 2x2 + 2x + 1 \(\ge0\)

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