Subfield: Algebra


Question: Let $n \geqslant 1$ be an integer, and let $x_{0}, x_{1}, \ldots, x_{n+1}$ be $n+2$ non-negative real numbers that satisfy $x_{i} x_{i+1}-x_{i-1}^{2} \geqslant 1$ for all $i=1,2, \ldots, n$. Show that

$$
x_{0}+x_{1}+\cdots+x_{n}+x_{n+1}>\left(\frac{2 n}{3}\right)^{3 / 2} .
$$

Solution:
Lemma 1.1. If $a, b, c$ are non-negative numbers such that $a b-c^{2} \geqslant 1$, then

$$
(a+2 b)^{2} \geqslant(b+2 c)^{2}+6
$$

Proof. $(a+2 b)^{2}-(b+2 c)^{2}=(a-b)^{2}+2(b-c)^{2}+6\left(a b-c^{2}\right) \geqslant 6$.

Lemma 1.2. $\sqrt{1}+\cdots+\sqrt{n}>\frac{2}{3} n^{3 / 2}$.

Proof. Bernoulli's inequality $(1+t)^{3 / 2}>1+\frac{3}{2} t$ for $0>t \geqslant-1$ (or, alternatively, a straightforward check) gives

$$
(k-1)^{3 / 2}=k^{3 / 2}\left(1-\frac{1}{k}\right)^{3 / 2}>k^{3 / 2}\left(1-\frac{3}{2 k}\right)=k^{3 / 2}-\frac{3}{2} \sqrt{k}
\tag{*}
$$

Summing up (*) over $k=1,2, \ldots, n$ yields

$$
0>n^{3 / 2}-\frac{3}{2}(\sqrt{1}+\cdots+\sqrt{n}) .
$$

Now put $y_{i}:=2 x_{i}+x_{i+1}$ for $i=0,1, \ldots, n$. We get $y_{0} \geqslant 0$ and $y_{i}^{2} \geqslant y_{i-1}^{2}+6$ for $i=1,2, \ldots, n$ by Lemma 1.1. Thus, an easy induction on $i$ gives $y_{i} \geqslant \sqrt{6 i}$. Using this estimate and Lemma 1.2 we get

$$
3\left(x_{0}+\ldots+x_{n+1}\right) \geqslant y_{1}+\ldots+y_{n} \geqslant \sqrt{6}(\sqrt{1}+\sqrt{2}+\ldots+\sqrt{n})>\sqrt{6} \cdot \frac{2}{3} n^{3 / 2}=3\left(\frac{2 n}{3}\right)^{3 / 2}
$$
Say that an index $i \in\{0,1, \ldots, n+1\}$ is $\operatorname{good}$, if $x_{i} \geqslant \sqrt{\frac{2}{3}} i$, otherwise call the index $i$ bad.

Lemma 2.1. There are no two consecutive bad indices.

Proof. Assume the contrary and consider two bad indices $j, j+1$ with minimal possible $j$. Since 0 is good, we get $j>0$, thus by minimality $j-1$ is a good index and we have

$$
\frac{2}{3} \sqrt{j(j+1)}>x_{j} x_{j+1} \geqslant x_{j-1}^{2}+1 \geqslant \frac{2}{3}(j-1)+1=\frac{2}{3} \cdot \frac{j+(j+1)}{2}
$$

that contradicts the AM-GM inequality for numbers $j$ and $j+1$.

Lemma 2.2. If an index $j \leqslant n-1$ is good, then

$$
x_{j+1}+x_{j+2} \geqslant \sqrt{\frac{2}{3}}(\sqrt{j+1}+\sqrt{j+2})
$$

Proof. We have

$$
x_{j+1}+x_{j+2} \geqslant 2 \sqrt{x_{j+1} x_{j+2}} \geqslant 2 \sqrt{x_{j}^{2}+1} \geqslant 2 \sqrt{\frac{2}{3} j+1} \geqslant \sqrt{\frac{2}{3} j+\frac{2}{3}}+\sqrt{\frac{2}{3} j+\frac{4}{3}},
$$

the last inequality follows from concavity of the square root function, or, alternatively, from the AM-QM inequality for the numbers $\sqrt{\frac{2}{3} j+\frac{2}{3}}$ and $\sqrt{\frac{2}{3} j+\frac{4}{3}}$.



Let $S_{i}=x_{1}+\ldots+x_{i}$ and $T_{i}=\sqrt{\frac{2}{3}}(\sqrt{1}+\ldots+\sqrt{i})$.

Lemma 2.3. If an index $i$ is good, then $S_{i} \geqslant T_{i}$.

Proof. Induction on $i$. The base case $i=0$ is clear. Assume that the claim holds for good indices less than $i$ and prove it for a good index $i>0$.

If $i-1$ is good, then by the inductive hypothesis we get $S_{i}=S_{i-1}+x_{i} \geqslant T_{i-1}+\sqrt{\frac{2}{3} i}=T_{i}$.

If $i-1$ is bad, then $i>1$, and $i-2$ is good by Lemma 2.1. Then using Lemma 2.2 and the inductive hypothesis we get

$$
S_{i}=S_{i-2}+x_{i-1}+x_{i} \geqslant T_{i-2}+\sqrt{\frac{2}{3}}(\sqrt{i-1}+\sqrt{i})=T_{i}
$$

Since either $n$ or $n+1$ is good by Lemma 2.1, Lemma 2.3 yields in both cases $S_{n+1} \geqslant T_{n}$