Proving that if $a^2+b^2=c^2$, then $a+b\ge c$. Hello, I'm trying to prove this statement.

Let a,b & c be three positive real numbers and if $a^2+b^2=c^2$ then $a+b\ge c$

Any help, please?

Suppose Not, which means that $a + b < c $
Square both sides
$(a + b)^2 < c^2 $
$a^2 + b^2 + 2ab < c^2 $
But by assumption $a^2 + b^2 = c^2$
Therefore,
$a^2 + b^2 + 2ab < a^2 + b^2 $
Which means
$2ab < 0 $ where a & b are positive numbers
Contradiction as no 2 positive multiplication is less than 0
therefore,
$a+b≥c $
