Le lundi 21 août 2023 à 23:06:21 UTC+2, DP a écrit :
Le dimanche 20 août 2023 à 18:39:42 UTC+2, [email protected]d a écrit :
Is there a simple way to show algebraically that:
b^2 >= (a - c)^2 AND a^2 >= b^2
implies:
c^2 >= (a + b)^2 OR c^2 >= (a - b)^2
where ">=" denotes "greater or equal"?
Martin.
Taking a = 2, b = 1, and c = -2 gives a counterexample as the first proposition is false, while the second true.
Dan
Sorry I misunderstood the problem, which is if the first proposition is true then the second is true as well.
To show that let x = a+b, and y = a-b. Then from the first inequation it comes after simplification
(c-x)(c-y) <= 0.
So there are two cases:
1) c <= x AND c >= y,
2) c >= x AND c <= y.
So either 1) is true and in this case
c >= y, that is c >= a-b, c^2 >= (a-b)^2,
or 2) is true and
c >= a+b, c^2 >= (a+b)^2.
Note that one could more completely say that
either c^2 >= (a-b)^2 AND c^2 <= (a+b)^2 OR c^2 >= (a+b)^2 AND c^2 <= (a-b)^2.
One could also keep the original inequations 1) and 2),
c >= a-b AND c <= a+b OR c >= a+b AND c <= a-b.
Also it seems I didn't use the second inequation a^2 >= b^2
which yields xy >= 0 (x and y have the same sign).
Dan
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