Prove or disprove $\sum\limits_{\mathrm{cyc}} \sqrt{a_1a_2} \ge \sum\limits_{\mathrm{cyc}} \sqrt[3]{a_1a_2a_3}$ for $0 \le a_1 \le \cdots \le a_n$

Question

Problem. Let $n\in \mathbb{N}_{\ge 3}$. Let $0 \le a_1 \le a_2 \le \cdots \le a_n$. Prove or disprove that $$\sqrt{a_1a_2} + \sqrt{a_2a_3} + \ldots + \sqrt{a_na_1} \ge \sqrt[3]{a_1a_2a_3} + \sqrt[3]{a_2a_3a_4} + \ldots + \sqrt[3]{a_na_1a_2}.$$

I saw this problem recently which was originally asked 11 years ago (a Problem Statement Question (PSQ) - which does not meet the current quality standard ).

Some observations:

i) By letting $a_i = x_i^6, \forall i$, the desired inequality becomes $$\sum\limits_{\mathrm{cyc}} x_1^3 x_2^3 \ge \sum\limits_{\mathrm{cyc}} x_1^2 x_2^2 x_3^2.$$

ii) @ivan investigated the case $n = 4, 5, 6$ and found that with the substitution $x_1 = y_1, x_2 = y_1 + y_2, \cdots, x_n = y_1 + y_2 + \cdots + y_n$, the polynomial $g(y_1, y_2, \cdots, y_n) := f(y_1, y_1 + y_2, \cdots, y_1 + y_2 + \cdots + y_n)$ has only one negative coefficient (this term is $- y_1^4 y_2 y_n$). It suffices to prove that this is the case for $n\ge 7$.

iii) Based on @ivan's observation, let $$F(x_1, x_2, \cdots, x_n) := \sum\limits_{\mathrm{cyc}} x_1^3 x_2^3 - \sum\limits_{\mathrm{cyc}} x_1^2 x_2^2 x_3^2 - (x_2 - x_1 - x_n + x_{n-1})^2 x_1^4.$$ Let $$G(y_1, y_2, \cdots, y_n) := F(y_1, y_1 + y_2, \cdots, y_1 + y_2 + \cdots + y_n).$$ It suffices to prove that all coefficients of $G$ are non-negative.

Answer 1

I'm leaving my previous approach at the end, but here's my much simpler proof of the inequality.

If $u\le v\le w$, we get $0\le(v-u)(w-v)=uv+vw-uw-v^2$: ie $uv+vw\ge uw+v^2$. We can apply this to $u,v,w=\sqrt{a_i},\sqrt{a_{i+1}},\sqrt{a_{i+2}}$ to get $$ \sqrt{a_ia_{i+1}}+\sqrt{a_{i+1}a_{i+2}}\ge \sqrt{a_ia_{i+2}}+a_{i+1} \quad\text{for}\quad i=1,\ldots,n-2 $$ which makes $$ \begin{align} 2\left(\sqrt{a_ia_{i+1}}+\sqrt{a_{i+1}a_{i+2}}\right)\ge & \sqrt{a_ia_{i+1}}+\sqrt{a_{i+1}a_{i+2}}+\sqrt{a_ia_{i+2}}+a_{i+1}\\ \ge & 3\,\sqrt[3]{a_ia_{i+1}a_{i+2}} + a_{i+1}. \end{align} $$ This makes $$ \sum_{i=1}^{n-2} \sqrt[3]{a_ia_{i+1}a_{i+2}}+\frac{a_{i+1}}3 \le \frac{2}{3}\left(\sqrt{a_1a_2}+\sqrt{a_{n-1}a_n}\right) +\frac43\left(\sqrt{a_2a_3}+\cdots+\sqrt{a_{n-2}a_{n-1}}\right). $$ Applying this, we get $$ \begin{align} \sum_\text{cyc}\sqrt{a_ia_{i+1}} \ge& \sqrt{a_1a_n}+\frac{\sqrt{a_1a_2}+\sqrt{a_{n-1}a_n}}{3} -\frac13\sum_{i=2}^{n-2}\sqrt{a_ia_{i+1}}\\ &+\sum_{i=1}^{n-2}\sqrt[3]{a_ia_{i+1}a_{i+2}} +\frac13\sum_{i=2}^{n-1}a_i. \end{align} $$ Using $D$ to denote the difference between the two sides, $$ \begin{align} D:=&\sum_\text{cyc}\sqrt{a_ia_{i+1}}-\sqrt[3]{a_ia_{i+1}a_{i+2}}\\ \ge& \sqrt{a_1a_n}+\frac{\sqrt{a_1a_2}+\sqrt{a_{n-1}a_n}}{3} -\frac13\sum_{i=2}^{n-2}\sqrt{a_ia_{i+1}}\\ &-\sqrt[3]{a_1a_2a_n}-\sqrt[3]{a_1a_{n-1}a_n} +\frac13\sum_{i=2}^{n-1}a_i \end{align} $$ to which we apply $\sqrt{a_ia_{i+1}}\le(a_i+a_{i+1})/2$ to obtain $$ D\ge \sqrt{a_1a_n}+\frac{\sqrt{a_1a_2}+\sqrt{a_{n-1}a_n}}{3} -\sqrt[3]{a_1a_2a_n}-\sqrt[3]{a_1a_{n-1}a_n} +\frac{a_2+a_{n-1}}6. $$ Next, using $x^sy^{1-s}\le sx+(1-s)y$ for $x,y\ge0$, $0\le s\le1$, entering $x=\sqrt{a_1a_2}, y=a_n, s=2/3$ makes $\sqrt[3]{a_1a_2a_n}\le\frac23\sqrt{a_1a_n}+\frac13\sqrt{a_2}$, etc. This makes $$ \begin{align} D\ge& \frac{\sqrt{a_1a_2}+\sqrt{a_{n-1}a_n}-\sqrt{a_1a_n}}{3} -\frac{a_2+a_{n-1}}{6}\\ =& \frac{\left(\sqrt{a_n}-\sqrt{a_2}\right) \left(\sqrt{a_{n-1}}-\sqrt{a_1}\right)}{3} -\frac{\left(\sqrt{a_{n-1}}-\sqrt{a_2}\right)^2}{6}\\ \ge& \frac{\left(\sqrt{a_{n-1}}-\sqrt{a_2}\right)^2}{6} \ge 0. \end{align} $$ And $D\ge0$ is exactly what we wanted to prove.

We also see from this that equality requires $a_1=a_2=a_{n-1}=a_n$ which means the $a_i$ are all equal.

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Here is my previous approach, somewhat too briefly described. I'll try to update it since the middle step was somewhat unusual. I have checked the steps by simulations (random $a_i$) and think the method should be sound, but there may be typos and mistakes even so.

Let $A(a_1,\ldots,a_n)=\sum_\text{cyc}\sqrt{a_ia_{i+1}}$ and $B(a_1,\ldots,a_n)=\sum_\text{cyc}\sqrt[3]{a_ia_{i+1}a_{i+2}}$. We want to show that $(A-B)(a_1,\ldots,a_n)\ge0$ whenever $0\le a_1\le\cdots\le a_n$.

The proof has three steps.

1. Show that $(A-B)(a_1,\ldots,a_n)\le(A-B)(a_1,\ldots,a_{i-1},a_{i+1},\ldots,a_n)$ for $3\le i\le n-2$.

2. Starting with $a_1,\ldots,a_n$, for any $\epsilon>0$ we can fill in the gaps between $a_i$ and $a_{i+1}$ forming a new sequence $a'_1\le\cdots\le a'_N$ containing $a_i$ as a subsequence, so that $a'_1=a_1$, $a'_2=a_2$, $a'_{n-1}=a_{N-1}$, $a'_{n}=a_{N}$, and $a'_{i+1}-a'_i<\epsilon$ for $2\le i<N-1$. Then $A(a)-B(a)\ge A(a')-B(a')$ by repeated use of the above inequality starting with $a'_1,\ldots,a'_N$ and removing elements one at a time to obtain $a_1,\ldots,a_n$.

3. Taking the limit as $\epsilon\rightarrow0$, we get a lower bound on $A(a)-B(a)\ge A(a')-B(a')$ which depends only on $a_1,a_2,a_{n-1},a_{n}$ which can be verified.

Proofs of the inequalities in 1. and 3. use the same approach as explained in the question: express inequality as polynomial in non-negative variables, and check for negative coefficients. I did this using the sympy Python package since I was already doing numerical simulations in Python to test hypotheses, but Sage is a good alternative.

Denoting $a=(a_1,\ldots,a_n)$ and $a^{(i)}=(a_1,\ldots,a_{i-1},a_{i+1},\ldots,a_n)$, $$ \begin{align} A(a)-A(a^{(i)}) =& \sqrt{a_{i-1}a_i}+\sqrt{a_ia_{i+1}}-\sqrt{a_{i-1}a_{i+1}}, \\ B(a)-B(a^{(i)}) =& \sqrt[3]{a_{i-2}a_{i-1}a_{i}}+\sqrt[3]{a_{i-1}a_{i}a_{i+1}} +\sqrt[3]{a_{i}a_{i+1}a_{i+2}} -\sqrt[3]{a_{i-2}a_{i-1}a_{i+1}}-\sqrt[3]{a_{i-1}a_{i+1}a_{i+2}}. \end{align} $$ If $3\le i\le n-2$, we can eliminate $a_{i-2}$ and $a_{i+2}$ by $$ \begin{align} B(a)-B(a^{(i)}) =& \sqrt[3]{a_{i-1}a_{i}a_{i+1}} -\sqrt[3]{a_{i-2}a_{i-1}}\cdot\left(\sqrt[3]{a_{i+1}}-\sqrt[3]{a_{i}}\right) +\sqrt[3]{a_{i+1}a_{i+2}}\cdot\left(\sqrt[3]{a_{i}}-\sqrt[3]{a_{i-1}}\right) \\ \ge& \sqrt[3]{a_{i-1}a_{i}a_{i+1}} -\sqrt[3]{a^2_{i-1}}\cdot\left(\sqrt[3]{a_{i+1}}-\sqrt[3]{a_{i}}\right) +\sqrt[3]{a^2_{i+1}}\cdot\left(\sqrt[3]{a_{i}}-\sqrt[3]{a_{i-1}}\right) \end{align} $$ If we let $a_{i-1}=x^6$, $a_i=(x+y)^6$, $a_{i+1}=(x+y+z)^6$ with $x,y,z\ge0$, we can express $$ \left[A(a^{(i)})-B(a^{(i)})\right] -\left[A(a)-B(a)\right] \ge P(x,y,z) $$ where $P(x,y,z)$ is a degree 6 polynomial with all non-negative coefficients. So, for $x,y,z\ge0$, we get $P(x,y,z)\ge0$, and so $$ A(a^{(i)})-B(a^{(i)}) \ge A(a)-B(a). $$

The original idea was to use induction starting with $n=4$, but that would require an inequality in the opposite direction: I confirmed this using simulations with random sequences.

Instead, I can insert extra values into the sequence, which reduces the difference $A-B$, and prove that $A-B\ge0$ even as $n$ increases. We can do so, except from the first and last two elements in the list which are not changed, and in such a way that we get a new sequence $a'_1\le\cdots\le a'_N$ as described in 2. Since $a'_{i+1}-a'_i<\epsilon=O(1/N)$ for $3\le i\le N-2$, we get $$ \sqrt{a'_ia'_{i+1}}-2\sqrt[3]{a'_ia'_{i+1}a'_{i+2}}+\sqrt{a'_{i+1}a'_{i+2}} = O(\epsilon^2) $$ as long as we take care to spread the $a'_i$ out evenly, eg $a'_i=\sqrt{a'_{i-1}a'_{i+1}}$ for all $a'_i$ not in the original sequence, and so the sum $$ \sum_{i=3}^{n-2} \sqrt{a'_ia'_{i+1}}-2\sqrt[3]{a'_ia'_{i+1}a'_{i+2}}+\sqrt{a'_{i+1}a'_{i+2}} \rightarrow 0\quad\text{as }\epsilon\rightarrow 0,\; N\rightarrow\infty. $$

Using the above limit, what remains of the sum $A(a')-B(a')$ becomes $$ \begin{align} A(a')-B(a')\rightarrow& \sqrt{a_1a_2}+\sqrt{a_1a_n}+\sqrt{a_{n-1}a_n}+\frac{a_2+a_{n-1}}{2} \\ &-\left( \sqrt[3]{a_1a_2^2}+\sqrt[3]{a_1a_2a_n} +\sqrt[3]{a_1a_{n-1}a_n}+\sqrt[3]{a_{n-1}^2a_n} \right) \end{align} $$ where we can enter $a_1=t^6$, $a_2=(t+u)^6$, $a_{n-1}=(t+u+v)^6$, $a_n=(t+u+v+w)^6$ with $t,u,v,w\ge0$ to obtain a polynomial $G(t,u,v,w)$ for the limit. Again, we expand and check the signs of the coefficients, and there is only one negative coefficient: $-t^4uw$. However, this is balanced out by coefficients $3t^4u^2$ and $3t^4w^2$, and the remaining terms are all non-negative. So $G(t,u,v,w)\ge0$ which proves that $$ A(a)-B(a) \ge \lim_{\epsilon\rightarrow0}A(a')-B(a') = G(t,u,v,w) \ge 0. $$

Answer 2

I'm adding a new solution following the approach used in the Mongolian inequality problem referred to by dezdichado. It hasn't been proof-read properly, so read with care.

It uses Karamata's inequality, which is a generalization of Jensen's inequality to sequences $x_1,\ldots,x_n$ and $y_1,\ldots,y_n$ where one majorizes the other.

We say that $(x_1,\ldots,x_n)$ majorizes $(y_1,\ldots,y_n)$ if, for $x^*_i$ and $y^*_i$ denoting the sequences sorted in decreasing order, we have $\sum_{i=1}^k x^*_i\ge \sum_{i=1}^k y^*_i$ for all $k=1,\dots,n$ and with equality for $k=n$: ie $\sum_{i=1}^n x_i=\sum_{i=1}^n y_i$. For any convex function $f$, Karamata's inequality states that $$\sum_{i=1}^n f\left(x^*_i\right) \ge \sum_{i=1}^n f\left(y^*_i\right).$$ We may note that, in terms of the sequences sorted in increasing order $\hat x_1\le\cdots\le\hat x_n$ and $\hat y_1\le\cdots\le\hat y_n$, the sequence $(x_1,\ldots,x_n)$ majorizes $y_1,\ldots,y_n$ if $\sum_{i=1}^k\hat x_i\le\sum_{i=1}^k\hat y_i$ for all $k$ with equality for $k=n$.

Let $a_i=e^{u_i}$ where we for the time being ignore the alternative of $a_i=0$ (eg solve by taking the limit $u_i\rightarrow-\infty$). Define sequences $x_i=\frac{u_i+u_{i+1}}{2}$ and $y_i=\frac{u_i+u_{i+1}+u_{i+2}}{3}$ with $i$ cyclic as before. Using $f(t)=e^t$, we have $$ \sum_i\sqrt{a_ia_{i+1}} = \sum_i f(x_i) \quad\text{and}\quad \sum_i\sqrt[3]{a_ia_{i+1}a_{i+2}} = \sum_i f(y_i). $$ If we can prove that $(x_1,\ldots,x_n)$ majorizes $(y_1,\ldots,y_n)$, then the inequality $\sum_i f(x_i)\ge\sum_i f(y_i)$ follows. The Mongolian inequality problem referred to above contains a proof of this, but here is a slightly different proof.

The sequences $x_i$ and $y_i$ start of being increasing, but with the last few term wrapping around: $$ x_1\le y_1\le x_2\le\cdots\le x_{n-2}\le y_{n-2}\le x_{n-1} \quad\text{and}\quad x_1\le y_n\le x_n\le y_{n-1}\le x_{n-1}. $$ We may express the sums of the sorted values using $(t)_+=\max(0,t)$ to insert the wrapped terms: $$ \sum_{i=1}^k\hat x_i = \sum_{i=1}^k x_i - \left(x_k-x_n\right)_+ \quad\text{and}\quad \sum_{i=1}^k\hat y_i = \sum_{i=1}^k y_i - \left(y_k-y_n\right)_+ - \left(y_{k-1}-y_{n-1}\right)_+. $$ We now want to show that the difference $$ D_k = \sum_{i=1}^k \left(\hat y_i-\hat x_i\right) = \sum_{i=1}^k \left(y_i-x_i\right) - \left(y_k-y_n\right)_+ + \left(x_k-x_n\right)_+ - \left(y_{k-1}-y_{n-1}\right)_+ $$ is non-negative. Note that the terms inside $(\cdot)_+$ are in order $y_k-y_n\ge x_k-x_n\ge y_{k-1}-y_{n-1}$.

If $y_k\le y_n$, the none of the $(\cdot)_+$ terms are positive, and so $6D_k = 2u_{k+2}+u_{k+1}-2u_2-u_1\ge0$.

If $y_{k-1}>y_{n-1}$, all the $(\cdot)_+$ are positive, and $6D_k = u_n+2u_{n-1}-u_k-2u_{k-1}\ge0$.

This leaves the case where $y_k>y_n$ but $y_{k-1}\le y_{n-1}$. In this case, $$ \begin{align} D_k =& \sum_{i=1}^k \left(y_i-x_i\right) - \left(y_k-y_n\right)_+ + \left(x_k-x_n\right)_+\\ \ge& \sum_{i=1}^k \left(y_i-x_i\right) - \left(y_k-y_n\right) + \frac{2}{3}\left(x_k-x_n\right) =\frac{u_{k+1}-u_1}{6} \ge 0. \end{align} $$ We can thus conclude that $(x_1,\ldots,x_n)$ majorizes $(y_1,\ldots,y_n)$, and the inequality therefore follows from Karamata's inequality.

Answer 3

Proof.

By AM-GM, we have $$\sqrt[3]{a_1a_2a_3} \le \frac{\sqrt{a_1a_2} + \sqrt{a_2a_3} + \sqrt{a_3a_1}}{3}, $$ $$\sqrt[3]{a_2a_3a_4} \le \frac{\sqrt{a_2a_3} + \sqrt{a_3a_4} + \sqrt{a_4a_2}}{3}, $$ $$\cdots \cdots \cdots \cdots,$$ $$\sqrt[3]{a_{n-2}a_{n-1}a_n} \le \frac{\sqrt{a_{n-2}a_{n-1}} + \sqrt{a_{n-1}a_n} + \sqrt{a_na_{n-2}}}{3},$$ $$\sqrt[3]{a_{n-1}a_n a_1} \le \frac{\sqrt{a_{n-1}a_n} + \sqrt{a_n a_1} + \sqrt{a_1 a_{n-1}}}{3},$$ and $$\sqrt[3]{a_na_1 a_2} \le \frac{\sqrt{a_n a_1} + \sqrt{a_n a_1} + a_2}{3}.$$ (Note: The last inequality is different from all the previous ones. )

Thus, we have \begin{align*} 3(\mathrm{LHS} - \mathrm{RHS}) &\ge 2\sqrt{a_1a_2} + \sum_{i=2}^{n-1} \sqrt{a_ia_{i+1}} - \sum_{i=1}^{n-2} \sqrt{a_i a_{i+2}} - \sqrt{a_1a_{n-1}} - a_2. \end{align*}

It suffices to prove that $$2\sqrt{a_1a_2} + \sum_{i=2}^{n-1} \sqrt{a_ia_{i+1}} - \sum_{i=1}^{n-2} \sqrt{a_i a_{i+2}} - \sqrt{a_1a_{n-1}} - a_2 \ge 0.$$

Letting $a_i = b_i^2, \forall i$, it suffices to prove that, for all $0\le b_1 \le b_2 \le \cdots \le b_n$, $$2b_1b_2 + \sum_{i=2}^{n-1} b_ib_{i+1} - \sum_{i=1}^{n-2} b_i b_{i+2} - b_1b_{n-1} - b_2^2 \ge 0. \tag{1}$$

We use the Mathematical Induction.

It is clearly true for $n = 3$ since (1) is written as $(b_3 - b_2)(b_2 - b_1) \ge 0$.

Assume that the statement is true for $n$ ($n\ge 3$).

For $n + 1$, we need to prove that, for all $0 \le b_1 \le b_2 \le \cdots \le b_n \le b_{n+1}$, $$2b_1b_2 + \sum_{i=2}^{n} b_ib_{i+1} - \sum_{i=1}^{n-1} b_i b_{i+2} - b_1b_{n} - b_2^2 \ge 0. \tag{2}$$ (2) is written as $$2b_1b_2 + \sum_{i=2}^{n-1} b_ib_{i+1} - \sum_{i=1}^{n-2} b_i b_{i+2} - b_1b_{n-1} - b_2^2$$ $$+ (b_n b_{n+1} - b_{n-1}b_{n+1} + b_1b_{n-1} - b_1 b_n) \ge 0$$ which is written as $$\left(2b_1b_2 + \sum_{i=2}^{n-1} b_ib_{i+1} - \sum_{i=1}^{n-2} b_i b_{i+2} - b_1b_{n-1} - b_2^2\right) + (b_n - b_{n-1})(b_{n+1}-b_1) \ge 0$$ which is true using the inductive hypothesis. Thus, the statement is true for $n+1$.

We are done.