solving the inequality with three parameters Is there any way to show that the following inequality holds for the given function with constraints?
$\frac{(a x + y)^{y+1}}{a x (a x + y + 1)^y}\geq 1$ for $0.5 \leq a \leq 1$, $x >0,y \geq 0$. It can be easily checked that it saturates the inequality for $\lim_{x \rightarrow \infty}$. 
Numerically, I've checked this for different range of $x,y$ and $a$. It always holds.. I'm not sure how I can prove this analytically!
Thanks!

Assume WLOG that $a=1$ (otherwise replace $ax \mapsto x\,$).
$$
\begin{align}
\frac{(x + y)^{y+1}}{x (x + y + 1)^y}\geq 1 \;\;&\iff \;\; \frac{x+y+1}{x} \, \left(\frac{x+y}{x+y+1}\right)^{y+1} \ge 1 \\
 &\iff\;\; \frac{x+y+1}{x}\,\left(1 - \frac{1}{x+y+1}\right)^{y+1} \ge 1
\end{align}
$$
The second factor on the LHS of the latter satisfies the conditions of Bernoulli's inequality:

for every real number $r \ge 1$ and real number $x \ge -1\,$: $\;(1+x)^r \ge 1+rx\,$.

Therefore 
$$\require{cancel}
\begin{align}
\frac{x+y+1}{x}\,\left(1 - \frac{1}{x+y+1}\right)^{y+1} &\ge \frac{x+y+1}{x}\left(1-\frac{y+1}{x+y+1}\right) \\
 &= \frac{\cancel{x+y+1}}{\bcancel{x}} \, \frac{\bcancel{x}}{\cancel{x+y+1}} \\
 &= 1
\end{align}
$$
Note that the condition $0.5 \leq a \leq 1$ was not used and is not required. It is enough that $a \gt 0$.
