Inequality for sides and height of right angle triangle

Question

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!

Answer 1

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.

Answer 2

The correct inequality is

$$(a+b)^2=(c+h)^2-xy<(c+h)^2$$

which implies $$a+b<c+h$$

Answer 3

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.

Answer 4

Perhaps easier, in the right triangle, once you note $ab=ch$, the smaller sum is when the terms are closer to each other, and clearly $c>a,b>h$, so $c+h>a+b$.