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 the denominator of the LHS is positive you are allowed to cross multiply. So it can be simplified as follows:-$$\begin{align}&(x+y+1)^2\le 4\left(\sqrt{x^2+x}+\sqrt{y^2+y}\right)^2\\&=>x^2+y^2+2xy+2x+2y+1\le 4x^2+4y^2+4x+4y+8\sqrt{(x^2+x)(y^2+y)}\\&=>2x^2+2y^2+(x-y)^2+2x+2y+8\sqrt{(x^2+x)(y^2+y)}-1\ge 0\end{align}$$ This is obviously true because $2(x+y)-1\ge 3$ as $x,y\ge 1$
