# Proof of this inequality

I have a finite sequence of positive numbers $(a_i)_1^n$ for which:

1. $a_1>a_n$,
2. $a_j\geq a_{j+1}\geq\cdots\geq a_n$ for some $j\in\{2,\ldots,n-1\}$,
3. $a_1>a_2>\cdots>a_{j-1}$,
4. $a_j\geq a_i$ for all $1\leq i\leq n$.

I conjecture that:

$$(a_n+a_2+\cdots+a_j)\left(\sum_{i=j+1}^{n-1}{\frac{a_i^2}{(a_1+\cdots+a_i)(a_n+a_2+\cdots + a_i)}}+\frac{a_1+a_n}{a_1+\cdots+a_n}\right) \geq (a_n+a_1+\cdots+a_j)\left(\sum_{i=j+1}^{n-1}{\frac{a_i^2}{(a_1+\cdots+a_i)(a_n+a_1+\cdots + a_i)}}+\frac{a_n}{a_1+\cdots+a_n}\right).$$

I have an unappealing brute-force proof when $n\in\{3,4,5\}$ but I can't prove it in general. I have tried calculus to no avail, and it doesn't seem like a good fit for any of the standard inequalities.Does this look even remotely similar to anything already done? I appreciate that it's a rather ugly inequality but some suggestions would be greatly appreciated!

The proof when $n=3$ is outlined below. By condition 2. we know that $j=2$ so that

\begin{align*} (a_3+a_2)\left(\frac{a_1+a_3}{a_1+a_2+a_3}\right)-(a_3+a_1+a_2)\left(\frac{a_3}{a_1+a_2+a_3}\right)=\frac{a_2(a_1-a_3)}{a_1+a_2+a_3}>0 \end{align*} where we have used condition 1, which states that $a_1>a_3$.The proofs when $n=4$ and $n=5$ are likewise, only uglier. When $n=5$ the trick is to find common denominators then simply pair each negative term with some larger positive term. It's horrible, but it works. Perhaps a general proof would involve a similar argument but more formalised?

If a full proof can't be found then I'd be happy for a proof in the special case where $j=2$ and $j=3$.

• Hint: Compare $a_1$ times smaller sum against the terms in the larger sum that come from having $a_1$ in the numerator of the rightmost fraction inside the sum (it's a bit messy to write down, but you should be able to get the desired inequality from those pieces, as the rest is shared by both sums). – Justin Benfield Apr 14 '16 at 3:52
• Condition (2) says "where $2\leq j\leq n-1$" -- is that supposed to mean "for all $j$" in that range? Or is there supposed to be some particular $j$ in that range for which the statement holds? Is Condition (3) then supposed to be stating that this particular $a_j$ is at least as large as any other $a_i$? (Otherwise, for which $j$ is Condition (3) supposed to hold?) – r.e.s. Apr 14 '16 at 4:56
• I will clarify. Thanks for pointing out the ambiguity. – Auslander Apr 14 '16 at 5:01
• So the inequality is only required to hold for some $j$ satisfying Conditions (2) and (3)? – r.e.s. Apr 14 '16 at 5:16
• Yes, that is right. The $j$ is determined by the sequence itself. Suppose we have the sequence $\{4,2,1,6,4,5,3\}$. Then $j$ could be $4$ or $5$. But in both instances, the left hand side is greater than the right hand side. – Auslander Apr 14 '16 at 5:33

Your conjecture fails for $n \ge 7$ (and maybe for some smaller $n$). The following results are from programming your inequality in Sage and testing a variety of patterned sequences.

1. First family of counterexamples

Let $S(a,b) = [a, a-1, a-2, ..., 1, b, b-1, b-2,..., 1]$, of length $a+b$, for positive integers $a,b$.

For any $a \ge 2$, there is some $b^* \gt a$ such that for any $b \ge b^*$, $S(a, b)$ is a counterexample. (The quantity $\ \text{LHS}-\text{RHS}\$ is a decreasing function of $b$, and falls below $0$ at $b=b^*$.)

Such counterexamples include $S(2, b\ge 11), S(3, b\ge 14), S(4, b\ge 16), S(5,b\ge 19)$, and so on. (I haven't determined a formula for $b^*$.)

The smallest counterexample of this form appears to be $$S(2,11) = [2, 1, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1]$$ for which
$$\text{LHS} = \frac{79582008868974649241}{15735265132809166560} = 5.0575... < 5.0608... = \frac{1185342437701}{234217526928} = \text{RHS}$$

2. Second (shorter) family of counterexamples

Another family of counterexamples is $[2, b, b, b, b, b, 1]$ with $b \ge 5$. The smallest counterexample of this form appears to be $[2, 5, 5, 5, 5, 5, 1]$, for which
$$\text{LHS} = \frac{3515219}{1225224} = 2.869... < 2.881... = \frac{30446902}{10567557} = \text{RHS}$$

Here's the core of the program (note the indexing adjustments due to Python lists being $0$-based):

def lhs(L,j):
n = len(L)
tot = (L + L[n-1])/sum(L)
for i in [j+1..n-2]: tot += L[i]^2 / ( sum(L[0:i+1])*( L[n-1] + sum(L[1:i+1]) ) )
return (L[n-1] + sum(L[1:j+1]))*tot

def rhs(L,j):
n = len(L)
tot = L[n-1]/sum(L)
for i in [j+1..n-2]: tot += L[i]^2 / ( sum(L[0:i+1])*( L[n-1] + sum(L[0:i+1]) ) )
return (L[n-1] + sum(L[0:j+1]))*tot

for b in [3..8]:
L = [2,b,b,b,b,b,1]; left = lhs(L,1); right = rhs(L,1)
print b, left.n(), right.n(), (left-right).n()

> 3 1.96695167577521 1.92603768780239 0.0409139879728115
> 4 2.41522132314971 2.40469223123685 0.0105290919128582
> 5 2.86904190580661 2.88116752055371 -0.0121256147470981
> 6 3.32586148147269 3.35536963036963 -0.0295081488969435
> 7 3.78448484495198 3.82769776396051 -0.0432129190085339
> 8 4.24427752155089 4.29855086580553 -0.0542733442546420

• I'm getting the same thing. This counterexample takes my paper one step forward and two steps back. Thanks for the step forward though! – Auslander Apr 15 '16 at 5:06

Edit: nvm, denominators of the first fraction in the sums are different.

Both sums share:

$(a_n+a_2+\cdots+a_j)\left(\sum_{i=j+1}^{n-1}{\frac{a_i^2}{(a_1+\cdots+a_i)(a_n+a_2+\cdots + a_i)}}+\frac{a_n}{a_1+\cdots+a_n}\right)$

If you subtract that out of the first sum, the following remains:

$(n-j-1)(a_n+a_2+\cdots+a_j)\left(\dfrac{a_1}{a_1+\cdots+a_i}\right)$ (why?)

Likewise for the 2nd sum:

$a_1\left(\sum_{i=j+1}^{n-1}{\frac{a_i^2}{(a_1+\cdots+a_i)(a_n+a_2+\cdots + a_i)}}+\frac{a_n}{a_1+\cdots+a_n}\right)$

remains. (why?)

• You write that both sums share: $$(a_n+a_2+\cdots+a_j)\left(\sum_{i=j+1}^{n-1}{\frac{a_i^2}{(a_1+\cdots+a_i)(a_n+a_2+\cdots + a_i)}}+\frac{a_n}{a_1+\cdots+a_n}\right).$$ I can't see how this is true - the denominator in each sum is slightly different. – Auslander Apr 14 '16 at 4:41
• You're right, didn't catch that...does make things more complicated. – Justin Benfield Apr 14 '16 at 10:41
• Yep. Wish it were that easy! But thanks for looking at it. I'll keep thinking. – Auslander Apr 14 '16 at 12:07