How to prove inequality $\;\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le2\;,\;\;\text{for}\;\;x,y\ge 1\;$ I have this great problem: to prove
$$\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le 2\;,\;\;\forall\;x,y\ge1\;?$$
Multiplicate by conjugated I get left side as
$$\frac{\sqrt{x^2+x}-\sqrt{y^2+y}}{x-y}\le 2$$
but I can't get any more, and I can't use derivatives or multiple limits since we didn't study these yet.
Thanks

Given $x,y\ge 1$.
$\therefore x<\sqrt{x^2+x}$ and $y<\sqrt{y^2+y}$ $\implies$ $x+y<\sqrt{x^2+x}+\sqrt{y^2+y}$
$$\frac{x+y+1}{x+y}>\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}} $$ and 
$\frac{x+y+1}{x+y}=1+\frac{1}{x+y}\le \frac{3}{2}$, because $x+y\ge2$

$$\therefore \frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}} <2 $$

