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?
 A: Assume $c\ge a+b$, then we have $c^2\ge (a+b)^2=a^2+2ab+b^2>a^2+b^2$. We arrived at a contradiction.
A: $a,b,c>0$ are the side lengths of a right triangle ($a,b$ the legs, $c$ the hypotenuse) iff $a^2+b^2=c^2$, by the (converse) Pythagorean theorem or by the cosine theorem. A right triangle is a triangle, hence its sides fulfill the triangle inequality $a+b > c$.
A: We have $c^2=a^2+b^2<a^2+2ab+b^2=(a+b)^2$, therefore, by taking squares and noting that $a,b,c$ are positive, we obtain $c<a+b$.
A: 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
$$
c^2=a^2+2ab+b^2-2ab(1+\tau)=(a+b)^2-2ab(1+\tau)<(a+b)^2
$$
A: For all $a,b \in \mathbb{Z}^+$ we have $a^2 + b^2 < (a+b)^2 \implies \sqrt{a^2+b^2} < a+b$. If $c^2$ is defined to be $c^2 = a^2 + b^2$, then we have $c < a+b$. This also explains why the triangle cannot be degenerate, i.e. $a = 0$ or $b = 0$, in order for statement to be true.
