Suppose a , b , and c are positive integers.
If a 2 + b 2 < c 2 , is it necessarily true that a + b < c ?
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NO.
As a counter-example, take 2,3,4. 2+3>4 but 2^2+3^2<4^2.
I couldn't think of a general explanation so I will give an example: (0.5^2)+(0.6^2) = 0.25 + 0.36 = 0.61< 1^2 = 1, but 0.5+0.6 = 1.1>1.
Therefore it is not always true that a^2 + b^2 < c^2.
While your argument can be easily modified, the question states that the variables are positive integers.
If a = b = 1 and c = 2 then 1 2 + 1 2 < 2 2 and 1 + 1 < 2 . Hence the answer is no.
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Consider a (non-degenerate) triangle with integer side lengths a , b , c and opposing angles A , B , C . Now if the triangle is obtuse with ∠ C > 9 0 ∘ then cos ( C ) < 0 , and so by the cosine law
c 2 = a 2 + b 2 − 2 a b cos ( C ) > a 2 + b 2 .
But by the triangle inequality a + b > c , so the given statement is not true, i.e., if a 2 + b 2 < c 2 it is not necessarily the case that a + b < c .