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Let $a_1,\dots,a_x$ be the positive divisors of $n$. Show that $\lceil n^{3/2}\rceil \geq a_1+\cdots+a_x$.

I've manage to transform that into $ \lceil a_x^{3/2}\rceil = \lceil (p_1^{q_1}p_2^{q_2}....p_k^{q_k})^{3/2} \rceil \ge (1+p_1+p_1^2+...+p_1^{q_1})(1+p_2+...+p_2^{q_2})....(1+p_k+...+p_k^{a_k}) $ sorry for my bad writing. I'm new here

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closed as off-topic by Sri-Amirthan Theivendran, Leucippus, Juniven, C. Falcon, Claude Leibovici Apr 15 '17 at 7:19

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    $\begingroup$ Welcome to Math.SE... Any thoughts and where you are stuck on this problem ? $\endgroup$ – Shailesh Apr 15 '17 at 0:01
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    $\begingroup$ Well, what if $n=2$? $\endgroup$ – Ned Apr 15 '17 at 0:17
  • $\begingroup$ @Shailesh I've manage to transform that into $ a_x^{3/2} = (p_1^{q_1}p_2^{q_2}....p_k^{q_k})^{3/2} >= (1+p_1+p_1^2+...p_1^{q_1})(1+p_2+...p_2^{q_2})....(1+p_k+...p_k^{a_k}) $ sorry for my bad writing. I'm new here $\endgroup$ – Hafizudeen Apr 15 '17 at 0:22
  • $\begingroup$ Edit your question and include that. Otherwise it might get closed. $\endgroup$ – Shailesh Apr 15 '17 at 0:24
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    $\begingroup$ You can do this by pairing up reciprocal divisors, and nothing that USUALLY, they sum to less than n. Since there are at most sqrt(n) pairs, you are basically done. The caveat is the pair (1,n), which sums to MORE than n. But, as long as there is another divisor, you'll be ok. So you can just worry about primes. All these things can be proved but that's up to you. $\endgroup$ – Steve D Apr 15 '17 at 0:25
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If $n = \prod_{p | n} p^{a_p(n)} $, then $\sigma(n) = \prod_{p | n} \sigma(p^{a_p(n)}) = \prod_{p | n} \dfrac{p^{a_p(n)+1}-1}{p-1} $.

Since both $n^{3/2}$ and $\sigma(n)$ are multiplicative functions, if we can show that $(p^{a_p(n)})^{3/2} \ge \dfrac{p^{a_p(n)+1}-1}{p-1} $, we are done.

Write this as $p^{3a/2} \ge \dfrac{p^{a+1}-1}{p-1} $.

Just to be a little more general, write $c$ for $3/2$, so this becomes $p^{ca} \ge \dfrac{p^{a+1}-1}{p-1} $, where $c > 1$.

Let's look at this for the small primes, since these cause the problems.

If $p=2$, this is $2^{ca} \ge 2^{a+1}-1 $. If $a=1$, this is $2^c \ge 3$ or $c \ge \log_2 3 \approx 1.585 $. If $a=2$, this is $4^{c} \ge 7 $ or $c \ge \frac{\log 7}{\log 4} \approx 1.403 $. The value of $c$ gets smaller for larger $a$, so $\log_2 3$ is the best we can do.

If $p=3$, this is $3^{ca} \ge \dfrac{3^{a+1}-1}{2} $, or $2\cdot 3^{ca} \ge 3^{a+1}-1 $. If we can show that $2\cdot 3^{ca} \ge 3^{a+1} $, we are done here.

This is $2\cdot 3^{ca-a-1} \ge 1 $ or $\ln 2+(a(c-1)-1)\ln 3 \ge 0$.

If $a \ge 2$, this is $\ln 2 +(2c-3)\ln 3 \ge 0$ which is certainly true for $c \ge 3/2$. This may be where the problem got the 3/2.

If $a=1$, this is $\ln 2+(c-2)\ln 3 \ge 0$ or $c \ge 2-\frac{\ln 2}{\ln 3} \approx 1.369 $. Again, $\frac32$ works.

We can now assume that $p \ge 5$.

We want $p^{ca} \ge \dfrac{p^{a+1}-1}{p-1} = \dfrac{p^{a+1}-p^a+p^a-1}{p-1} = p^a+\dfrac{p^a-1}{p-1} $, so if $p^{ca} \ge 2p^a$, we are done.

This is $p^{a(c-1)} \ge 2$. Since $p \ge 5$ and $a \ge 1$, this is true when $5^{c-1} \ge 2 $ or $c \ge 1+\frac{\ln 2}{\ln 5} \approx 1.43 $, so the $3/2$ works.

So $\frac32$ works for everything except $n=2$, in which case we have to use $\log_2 3 \approx 1.585 $.

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    $\begingroup$ Proving $p^{3a/2} \ge \dfrac{p^{a+1}-1}{p-1}$. If $a \ge2$, $p^{3a/2} \ge p^{a+1}>\dfrac{p^{a+1}-1}{p-1}$. If $a=1$ and $p \ge 3$, $p^{1.5}=\sqrt{p}p\ge \sqrt{2}p>p+1=\dfrac{p^{a+1}-1}{p-1}$. If $a=1$ and $p=2$, counterexample. $\endgroup$ – didgogns Apr 15 '17 at 2:04
  • $\begingroup$ That's what I said: 3/2 for everything except n=2. $\endgroup$ – marty cohen Apr 15 '17 at 3:34

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