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?
 A: $(a+b)^2=a^2+b^2+2ab\geq a^2+b^2= c^2\rightarrow (a+b)^2\geq c^2\rightarrow a+b\geq c$
A: Hints:
(1) If for positive real numbers $\;a,b,c\;$ we have that $\;a^2+b^2=c^2\;$ , then there exists a right triangle with legs $\;a,b\;$ and hypotenuse $\;c\;$
(2) In Euclidean Geometry : the sum of the lengths of any two sides of any triangle is greater than the length of the third side.
A: $ a  =\! \sqrt{c^2\!-b^2} =\! \sqrt{{(c\!-\!b)(c\!+\!b)}}\, >\, c\!-\!b\,\ $  by $\,\ c\!-\!b < c\!+\!b$
A: Hint: Add $2ab$ and see what gives.
A: Consider $(a+b)^2=a^2+2ab+b^2=c^2+2ab$
Since $2ab>0$ as $a,b>0$, we have 
$$(a+b)^2 > c^2$$
Square rooting both sides gives the desired result.
A: HINT: Consider $a,b,c$ as sides of a right triangle.
A: 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 $
