perplexus.info :: Just Math : At most

At most - equality (Posted on 2017-03-06)

Prove:
For any real numbers a,b,c
ab + bc + ac does not exceed a^2+b^2+c^2

No Solution Yet Submitted by Ady TZIDON

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solution

Comment 1 of 1

consider 2(a^2 + b^2 + c^2) - 2(ab + bc + ac)

= (a^2 - 2ab +b^2) + (a^2 - 2ac + c^2) + (b^2 - 2bc + c^2)

= (a-b)^2 + (a-c)^2 + (b-c)^2

>= 0 (since each square term is itself >=0, so is their sum.)

well, if

2(a^2 + b^2 + c^2) - 2(ab + bc + ac) >= 0 then

(a^2 + b^2 + c^2) - (ab + bc + ac) >= 0 and

(a^2 + b^2 + c^2) >= (ab + bc + ac)

Posted by Paul on 2017-03-06 08:45:57

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