<s>[INST] 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
 [/INST] 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 $x,y\ge 1$
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