Set $ABC$ be a right traingle such that $BC=a,\ CA=b,\ AB=c.$ and $\angle{BCA=90^\circ}.$ Let $\angle {ABC} =\theta ,$ we have $a=c\cos \theta ,\ b=c\sin \theta , h=a\sin \theta =c\sin \theta \cos \theta .$ Thus $\frac{c+h}{a+b}=\frac{c(1+\sin \theta \cos \theta)}{c(\sin \theta +\cos \theta)}.$ Let $x=\sin \theta +\cos \theta,$ by $(\sin \theta +\cos \theta)^2=1+2\sin \theta \cos \theta ,$ we have $\sin \theta \cos \theta =\frac{x^2-1}{2}.$ so we can rewrite this equation as $\frac{c+h}{a+b}=\frac{1}{2}\left(x+\frac{1}{x}\right).$ Now the range of $x$ is $1< x\leq \sqrt{2},$ because we can write $x=\sqrt{2}\sin \left(\theta +\frac{\pi}{4}\right),$ from $0<\theta <\frac{\pi}{2}\Longrightarrow \frac{\pi}{4} <\theta +\frac{\pi}{4}<\frac{3}{4}\pi ,$ yielding $1<x \leq \sqrt{2}.$ Since the graph of $y=\frac{1}{2}\left(x+\frac{1}{x}\right)$ is an increasing function on the range of $x.$ Therefore $\frac{c+h}{a+b}$ is maximized when $x=\sqrt{2}\Longleftrightarrow \theta =\frac{\pi}{4},$ the desired maximum value is $\boxed{\frac{3\sqrt{2}}{4}}.$

Then the smaller sides be a,b then the expression becomes c+h/a+b = (a^2+b^2+ab)/(a+b)(sqrt(a^2+b^2))). Then, be a simple substitution, this reduces to: (k^2+k+1)/(k+1)(sqrt(k^2+1)). This does not exceed 3/sqrt8. To prove this inequality, aquare both sides and apply AMGM in six variables or weighted AM-GM, in which case we are done. Bomb

See http://www.mathlinks.ro/Forum/viewtopic.php?t=79510

Sailor, I didn't know. Valentin Vornicu asked me to post the problems.

We will prove that the maximum value is $\frac{3\sqrt{2}}{4}$. The inequality is equivalent to: $\frac{a^2+b^2+ab}{\sqrt{a^2+b^2}(a+b)}\leqslant\frac{3\sqrt{2}}{4}$ After squaring both parts we have: $(a^2+b^2+ab)^{2}\leqslant (a^{2}+b^2)(a+b)^{2}\cdot \frac{9}{8} \Longleftrightarrow $ $8(a^4+b^4)+24(ab)^{2}+16ab(a^2+b^2)\leqslant 9(a^2+b^2)(a+b)^{2}= \\ =9(a^4+b^4)+18(ab)^{2}+18a^3b+18b^3a$. After simplifying we obtain:

$a^4+b^4+2(a^3b+b^3a)\geqslant 6(ab)^2$ which is true from AM-GM