<s>[INST] Proving that if $c^2 = a^2 + b^2$ then $c < a + b$ I am having some trouble in proving a conjecture that occurred to me some time ago, based on the Pythagorean theorem.
If, for a non-degenerate triangle, $$c^2 = a^2 + b^2$$
Then can the following be proven?
$$c < a + b$$
Is this statement always true?
 [/INST] I guess your doubt is: 

does a triple $(a,b,c)$ of positive numbers with $c^2=a^2+b^2$ also satisfy $c<a+b$, so that the three numbers are the sides of a triangle?

In particular, does a Pythagorean triple define a (right) triangle?
Indeed, the other two conditions, namely $a<b+c$ and $b<a+c$, are obviously satisfied because from $c^2=a^2+b^2$ we immediately get $c>a$ and $c>b$.
Also the third condition holds:
$$
c^2=a^2+b^2<a^2+2ab+b^2=(a+b)^2
$$
Note where the positivity assumption is applied.
More generally, suppose $c^2=a^2+b^2-2ab\tau$, where $-1<\tau<1$ (the cosine law should come to mind). Then
$$
c^2=a^2+2ab+b^2-2ab(1+\tau)=(a+b)^2-2ab(1+\tau)<(a+b)^2
$$
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