Inequality for sides and height of right angle triangle 
Someone recently posed the question to me for the above, is c+h or a+b greater, without originally the x and y lengths. I used this method: (mainly pythagorus)
$a^2+b^2=c^2=(x+y)^2=x^2+y^2+2xy$
$a^2=x^2+h^2$ and $b^2=y^2+h^2$
therefore $x^2+h^2+y^2+h^2=x^2+y^2+2xy$
$x^2+y^2+2h^2=x^2+y^2+2xy$
so $2h^2=2xy$ and $$xy=h^2$$
also $Area={ab\over 2}={ch\over 2}$ so $$ab=ch$$
$(a+b)^2=a^2+b^2+2ab$
$(c+h)^2=c^2+h^2+2ch=a^2+b^2+xy+2ab$ 
therefore $$(a+b)^2+xy=(c+h)^2$$
so $$c+h>a+b$$
I feel like I have made a mistake somewhere, is this incorrectly generalised? And is there an easier way to show the inequality. Also, if this is correct can it be expanded to non right angle triangle, I tried to do this using trig but was pretty much going in circles, thanks!

I think we get $$a+b<c+h$$
Squaring we obtain
$$a^2+b^2+2ab<c^2+h^2+2hc$$
thus we have
$$2ab<h^2+2hc$$
$$2hc<h^2+2hc$$
because $ab=ch$ (area formulas) and we get $$h^2>0,$$ which is true.
