# Upper bound of a sum

It is given that $\sum_{i=1}^{n} x_i^3 \leq C$

Show that there exist a positive real number $K$ which satisfy the following inequality:

$\sum_{i=1}^{n}\frac{x_i}{a x_i + b} \leq K$

Where $a,b,C >0$ and $x_i>0 ; \forall i\in\{1,2,3,...,n\}$

I tried to solve this but don't have any idea where to start! I strongly doubt if this inequality is true. Is it possible to disprove in case it's untrue.

• You tagged the question "contest math", is it because you have a source? – Vincenzo Oliva Aug 8 '15 at 20:08
• It's from a regional Olympiad according to my friend. But I don't have the exact source and that's why I am removing tag from 'contest-math'. – Jenn Aug 8 '15 at 20:13
• Is $K$ allowed to depend on $n$? I assume not, right? – JimmyK4542 Aug 8 '15 at 20:14
• Does it depend on $a,b$? – Tryss Aug 8 '15 at 20:15
• I think $K$ is a constant which depends on $a,b$ and $C$. – Jenn Aug 8 '15 at 20:25

$$f(x) = \frac{x^{1/3}}{a x^{1/3}+b}$$ is a concave function on $\mathbb{R}^+$, hence the claim follows from Jensen's inequality. It is enough to take $K$ as: $$K=n \frac{(C/n)^{1/3}}{b+a(C/n)^{1/3}}.$$ $K$ has to depend on $n$, since otherwise we may take $x_i\approx\frac{1}{i^{2/3}}$ and in such a case $$\sum_{n\geq 1}\frac{x_i}{ax_i+b}$$ is not a convergent series, as Vincenzo Oliva pointed in the comments.
• Thanks Jack. It's a great help. But is it possible to find $K$ from given constraint. – Jenn Aug 8 '15 at 20:31
• @Ria: $$K = 3\frac{(C/3)^{1/3}}{a(C/3)^{1/3}+b}.$$ – Jack D'Aurizio Aug 8 '15 at 20:39
• Almost surely I'm missing something, but consider the following. Let $x_i=\displaystyle\frac{1}{i^{2/3}}$. Then $$\sum_{i=1}^n x_{i}^3 = \sum_{i=1}^n \frac{1}{i^2}<\frac{\pi^2}{6}=C.$$ But, with e.g. $a=b=1$, $$\sum_{i=1}^\infty\frac{1/i^{2/3}}{1/i^{2/3}+1}=\sum_{i=1}^\infty \frac{1}{i^{2/3}+1}=\infty,$$ so for large enough $n$ $$\sum_{i=1}^n \frac{1}{i^{2/3}+1}>3\frac{(\pi^2/18)^{1/3}}{(\pi^2/18)^{1/3}+1}=K.$$ What am I missing? – Vincenzo Oliva Aug 8 '15 at 21:53
• @VincenzoOliva: for some unknown reason I assumed $n=3$. No, well, given a finite $n$, Jensen's inequality gives that our sum is bounded by $$n \frac{(C/n)^{1/3}}{b+a(C/n)^{1/3}}.$$ – Jack D'Aurizio Aug 8 '15 at 23:27
• Now I see. Indeed despite having not that much of a practice with Jensen's inequality,I did expect to see $n$. +1, by the way. – Vincenzo Oliva Aug 9 '15 at 7:32