<s>[INST] 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!
 [/INST] It is not true for all triangles. 
From the law of cosines, $c^2 = b^2+ a^2 - 2ab \cos \gamma$, we can see that it is true when the triangle is obtuse.
On the other hand, it is not true for a triangle where all sides are equal. 
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