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

Since $x\ge0$, $\sqrt{x^2+x}\ge x$, and since $x\ge1$, $\sqrt{x^2+x}\ge \sqrt2$. Thus,
$$
\frac{x+y}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le1
$$
and
$$
\frac1{\sqrt{x^2+x}+\sqrt{y^2+y}}\le\frac1{2\sqrt2}
$$
we have
$$
\frac{x+y+1}{\sqrt{x^2+x}+\sqrt{y^2+y}}\le1+\frac1{2\sqrt2}\lt2
$$
