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Let $a_n \geq 0$ for each $n$. I have the relation $$a_n^2 \leq C_1a_na_{n-1} + C_2a_{n-1} + C_3a_n$$ where the constants $C_i > 0$ don't depend on $n$ and in fact they can be made as small as necessary.

I am wondering if I find a bound on $a_n$ that is independent of $n$ (it can depend on $a_0$ or $a_1$). Is it ppossible? Eg. I want $$a_n \leq C.$$

If $C_2=C_3=0$ then this holds if we take $C_1 \leq 1$. Otherwise I don't see it.

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    No. For example, $a_n=n$ solves $$a_n^2=a_na_{n-1}+a_n\leqslant 2a_na_{n-1}+3a_n+4a_{n-1}$$ – 2017-02-08
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    Right, but shouldn't it help that I can make the constants as small as I want? – 2017-02-08
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    Assume that $$a_n^2\leqslant\tfrac12a_na_{n-1}+\tfrac12a_n+\tfrac12a_{n-1}$$ for every $n>N$, then either $a_n\leqslant a_{n-1}$, or $a_n>a_{n-1}$ and then $$a_n\leqslant\tfrac12a_{n-1}+\tfrac12+\tfrac12a_{n-1}a_n^{-1}\leqslant\tfrac12a_{n-1}+\tfrac12+\tfrac12=\tfrac12a_{n-1}+1$$ Thus, for every $n>N$, $$a_n\leqslant\max\{a_{n-1},\tfrac12a_{n-1}+1\}$$ which implies that $$a_n\leqslant\max\{a_N,2\}$$ – 2017-02-08
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    Is it safe to assume that the recurrence inequality is valid only for large $n$, or precisely, for $n \geq N$ where $N$ depends on $C_i$'s and $a_1$? – 2017-02-08
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    @SangchulLee Um I'm not sure what you mean. The recurrence relation holds for all $n$ and the constants $C_i$ are basically given "data" that I'm allowed to choose and they don't influence the $a_n$. I hope that makes sense.. – 2017-02-08
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    Oh, I thought that $(a_n)$ is fixed and $C_i$ can be chosen arbitrarily small. In that case, assume that $r < 1$ and $C_i < r$ for $i = 1, 2, 3$. Then $a_n^2 \leq r(a_n a_{n-1} + a_n + a_{n-1})$ and this is equivalent to $$ a_n \leq r a_{n-1} + (1+r)\frac{a_n}{a_n+1}. $$ This implies $a_n \leq r a_{n-1} + 1 + r$. It is routine to prove that such $(a_n)$ is bounded. – 2017-02-08
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    I don't understand the question. Other than the zero sequence, surely there is no fixed sequence which satisfies the given inequality for all choices of $C_1,C_2,C_3 > 0$. – 2017-02-08
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    @Did thanks for the answer – 2017-02-08
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    @SangchulLee Thanks for the calculation there – 2017-02-08
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    To be completely explicit, my first comment shows that some unbounded sequences satisfy the desired property for some fixed $C_i$s (namely, $C_1=C_3=1$, $C_2=0$). My second comment shows that if a sequence satisfies the desired property, for $n$ large enough, with each $C_i\leqslant\frac12$, then it is bounded. In particular, if, as the OP seems to ask, if $$a_n^2\leqslant C_1(n)a_na_{n-1}+C_2(n)a_{n-1}+C_3(n)a_n$$ for every $n$, for some sequences $(C_i(n))$ all converging to zero, then $C_i(n)\leqslant\frac12$ for $n$ large enough, hence our result shows that $(a_n)$ is bounded. – 2017-02-08
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    More generally, I suspect that the conditions $$\limsup C_1(n)<1\qquad\limsup C_2(n)+C_3(n)<\infty$$ suffice to ensure that $(a_n)$ is bounded. – 2017-02-08

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Suppose that a sequence $a_0,a_1,a_2, ... $ of nonnegative real numbers satisfies

$$a_n^2 \leq C_1a_na_{n-1} + C_2a_{n-1} + C_3a_n$$

for all positive integers $n$ and all choices of positive real constants $C_1,C_2,C_3$.

Solving the recurrence for $a_{n-1}$ yields

$$a_{n-1} \ge \frac{a_n(a_n - C_3)}{C_1a_n + C_2}$$

For fixed $C_3 > 0$, if we let $C_1,C_2$ approach $0$ from above, the RHS will approach infinity unless $a_n \le C_3$.

Thus, $a_n \le C_3$ for all $n > 0$.

Since $C_3$ can be made arbitrarily small, it follows that $a_n = 0$ for all $n > 0$.

Thus, if I've understood the problem correctly, the only such sequences are those for which $a_0 \ge 0$ and $a_n = 0$ for all $n > 0$.

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    Thanks for your answer. I think what you proved is it;s possible to choose the constants so that the only sequence is the zero sequence which satisfies the recurrence. But isn't it possible to find constants $C_i$ so that the desired estimate is true and the sequence isn't zero? – 2017-02-08
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    Perhaps I don't understand the specification. Is the sequence fixed _before_ $C_1,C_2,C_3$ are chosen? That's the interpretation I used for my answer. – 2017-02-08
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    Suppose either a) the sequence depends on the constants or b) the sequence is fixed before the constants are chosen. I guess we can conclude that if a) then the desired bound is possible, and if b) then the only solution is zero. Is that right? – 2017-02-08
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    I'll agree with (b) (except that there's no bound on $a_0$). That was what I proved in my posted answer. As far as (a), wasn't a counterexample already posted in the comment by "Did"? – 2017-02-08
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    For (a), Did fixes the sequence first and then the constants (I think...) and Did's second comment proves a bound for any sequence such that the recurrence holds with those constants, doesn't it? – 2017-02-08
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    Well, perhaps you should ask Did, but as I understand his comment, if you fix the constants as $C1,C2,C3 = 2,3,4$, you can't prove a bound since for those constants, the sequence $a_n = n$ satisfies the inequality. So no, he's not fixing the sequence first. He's fixing $C_1,C_2,C_3 = 2,3,4$. Fixing the sequence first is what _I_ did in my solution. – 2017-02-08
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    @quasi Please se my last comment on main for what (I think) the OP really asks. – 2017-02-08
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    @Did: Ok, that makes sense. I'll delete my answer. – 2017-02-08
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    In other words, the intended question was perhaps this: For which triples $C_1,C_2,C_3 > 0$ does the given inequality force the sequence to be bounded, and, for those triples $C_1,C_2,C_3$ where boundedness is forced, can a bound $C$ be expressed in terms $C_1,C_2,C_3$ and possibly some initial terms of the sequence? – 2017-02-08
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    @quasi I think that's the best way to phrase it. But please consider leaving your answer undeleted as it is useful – 2017-02-08