Triangle Inequality Extension

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 $\angle C \gt 90^{\circ}$ then $\cos(C) \lt 0$ , and so by the cosine law

$c^{2} = a^{2} + b^{2} - 2ab\cos(C) \gt a^{2} + b^{2}$ .

But by the triangle inequality $a + b \gt c$ , so the given statement is not true, i.e., if $a^{2} + b^{2} \lt c^{2}$ it is not necessarily the case that $a + b \lt c$ .

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.

If $a = b = 1$ and $c = 2$ then $1^2 + 1^2 < 2^2$ and $1 + 1 \not< 2$ . Hence the answer is no.