# 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!

• but for $x=0$ is your term not defined – Dr. Sonnhard Graubner Jul 16 '17 at 22:37
• Let me edit my question and exclude x=0 case. I actually don't need that particular case in my problem.. – kphy Jul 16 '17 at 22:41

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 $a \gt 0$.

• nice proof and a good idea! – Dr. Sonnhard Graubner Jul 16 '17 at 22:45
• @Dr.SonnhardGraubner Thanks. As it happens, I do think that Bernoulli's is underrated as a general problem solving technique (not just for proving other well known inequalities). – dxiv Jul 16 '17 at 22:48
• That's great! Thank you @dxiv! – kphy Jul 16 '17 at 22:53
• Hey @dxiv.. can you check my new question? – kphy Jul 17 '17 at 7:02