Kvant Math Problem 1150
For $n=3$ the inequality becomes
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m09s
Source on kvant.digital
Problem
Prove that for any positive $a_1$, $a_2$, $a_3$, $\ldots$, $a_n$, the inequality $$\dfrac{(a_1+a_2+\ldots+a_n)^2}{2(a_1^2+a_2^2+\ldots+a_n^2)}\le\dfrac{a_1}{a_2+a_3}+\dfrac{a_2}{a_3+a_4}+\ldots+\dfrac{a_n}{a_1+a_2}.$$ holds.
E. G. Moiseev, 10th-grade student
Exploration
For $n=3$ the inequality becomes
$$\frac{(a_1+a_2+a_3)^2}{2(a_1^2+a_2^2+a_3^2)} \le \frac{a_1}{a_2+a_3}+\frac{a_2}{a_3+a_1}+\frac{a_3}{a_1+a_2}.$$
The right-hand side is the expression from Nesbitt's inequality, whose lower bound is $3/2$. The left-hand side is at most
$$\frac{3(a_1^2+a_2^2+a_3^2)}{2(a_1^2+a_2^2+a_3^2)} =\frac32$$
by Cauchy. Hence the statement is true for $n=3$.
This suggests separating the problem into two independent estimates. The left-hand side contains $(\sum a_i)^2$ and $\sum a_i^2$, so Cauchy's inequality immediately gives
$$\frac{(\sum a_i)^2}{2\sum a_i^2}\le \frac n2.$$
Thus it would suffice to prove
$$\sum_{i=1}^n \frac{a_i}{a_{i+1}+a_{i+2}}\ge \frac n2,$$
where indices are taken cyclically.
Trying small cases, for $n=4$ and $a_i=1$ the sum equals $2$, exactly $n/2$. This makes $\frac n2$ a natural target.
The expression
$$\frac{a_i}{a_{i+1}+a_{i+2}}$$
has the form suitable for Titu's lemma:
$$\sum \frac{a_i^2}{a_i(a_{i+1}+a_{i+2})} \ge \frac{(a_1+\cdots+a_n)^2} {\sum a_i(a_{i+1}+a_{i+2})}.$$
Hence everything reduces to estimating the denominator. We have
$$\sum a_i(a_{i+1}+a_{i+2}) = \sum a_i a_{i+1}+\sum a_i a_{i+2}.$$
Each of these sums is at most $\sum a_i^2$ by
$$2\sum a_i a_{i+k}\le 2\sum a_i^2.$$
Therefore
$$\sum a_i(a_{i+1}+a_{i+2}) \le 2\sum a_i^2.$$
Combining this with Titu's lemma gives
$$\sum \frac{a_i}{a_{i+1}+a_{i+2}} \ge \frac{(\sum a_i)^2}{2\sum a_i^2},$$
which is exactly the desired inequality.
The step most likely to hide an error is the estimate
$$\sum a_i a_{i+k}\le \sum a_i^2.$$
It must be justified carefully for cyclic indices.
Problem Understanding
We must prove that for positive numbers $a_1,\dots,a_n$,
$$\frac{(a_1+\cdots+a_n)^2}{2(a_1^2+\cdots+a_n^2)} \le \sum_{i=1}^{n} \frac{a_i}{a_{i+1}+a_{i+2}},$$
where the indices are taken cyclically.
This is a Type B problem. The goal is to establish a universal inequality.
The core difficulty is obtaining a lower bound for the cyclic sum on the right that naturally involves the quantities $(\sum a_i)^2$ and $\sum a_i^2$ appearing on the left. The appropriate tool is Engel's form of Cauchy's inequality, after which the denominator can be bounded by $\sum a_i^2$ through elementary quadratic estimates.
Proof Architecture
First lemma: for positive numbers $x_i,y_i$,
$$\sum \frac{x_i^2}{y_i}\ge \frac{(\sum x_i)^2}{\sum y_i}.$$
This is Engel's form of Cauchy's inequality.
Second lemma: for every fixed cyclic shift $k$,
$$\sum_{i=1}^n a_i a_{i+k}\le \sum_{i=1}^n a_i^2.$$
This follows from the inequality $2xy\le x^2+y^2$ applied termwise and summed.
Third lemma:
$$\sum_{i=1}^n a_i(a_{i+1}+a_{i+2}) \le 2\sum_{i=1}^n a_i^2.$$
This is obtained by applying the second lemma to the shifts $1$ and $2$.
The hardest step is the second lemma, because the cyclic indexing must be handled correctly when summing the inequalities.
Solution
Let
$$S=\sum_{i=1}^{n}\frac{a_i}{a_{i+1}+a_{i+2}},$$
with indices understood modulo $n$.
Applying Engel's form of Cauchy's inequality to
$$\frac{a_i}{a_{i+1}+a_{i+2}} = \frac{a_i^2}{a_i(a_{i+1}+a_{i+2})},$$
we obtain
$$S = \sum_{i=1}^{n} \frac{a_i^2}{a_i(a_{i+1}+a_{i+2})} \ge \frac{(a_1+\cdots+a_n)^2} {\sum_{i=1}^{n}a_i(a_{i+1}+a_{i+2})}.$$
It remains to estimate the denominator.
For any integer $k$, the inequality
$$2a_i a_{i+k}\le a_i^2+a_{i+k}^2$$
holds. Summing over $i$ gives
$$2\sum_{i=1}^{n} a_i a_{i+k} \le \sum_{i=1}^{n} a_i^2+\sum_{i=1}^{n} a_{i+k}^2.$$
Because the indices are cyclic, the second sum is merely a reordering of the first:
$$\sum_{i=1}^{n} a_{i+k}^2=\sum_{i=1}^{n} a_i^2.$$
Hence
$$\sum_{i=1}^{n} a_i a_{i+k} \le \sum_{i=1}^{n} a_i^2.$$
Applying this with $k=1$ and $k=2$ yields
$$\sum_{i=1}^{n} a_i a_{i+1} \le \sum_{i=1}^{n} a_i^2, \qquad \sum_{i=1}^{n} a_i a_{i+2} \le \sum_{i=1}^{n} a_i^2.$$
Therefore
$$\sum_{i=1}^{n}a_i(a_{i+1}+a_{i+2}) = \sum_{i=1}^{n} a_i a_{i+1} + \sum_{i=1}^{n} a_i a_{i+2} \le 2\sum_{i=1}^{n} a_i^2.$$
Substituting this estimate into the bound obtained from Engel's inequality, we get
$$S \ge \frac{(a_1+\cdots+a_n)^2} {2(a_1^2+\cdots+a_n^2)}.$$
This is exactly the required inequality.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the application of Engel's inequality. Writing
$$\frac{a_i}{a_{i+1}+a_{i+2}} = \frac{a_i^2}{a_i(a_{i+1}+a_{i+2})}$$
puts the sum in the standard form
$$\sum \frac{x_i^2}{y_i}.$$
Using Engel's inequality then gives
$$\sum \frac{x_i^2}{y_i} \ge \frac{(\sum x_i)^2}{\sum y_i}.$$
Taking $x_i=a_i$ and $y_i=a_i(a_{i+1}+a_{i+2})$ produces the claimed bound.
The second delicate step is proving
$$\sum a_i a_{i+k}\le \sum a_i^2.$$
Starting from
$$2a_i a_{i+k}\le a_i^2+a_{i+k}^2$$
and summing gives
$$2\sum a_i a_{i+k} \le \sum a_i^2+\sum a_{i+k}^2.$$
The cyclic indexing is crucial here. Because $i\mapsto i+k$ is a permutation of the residue classes modulo $n$,
$$\sum a_{i+k}^2=\sum a_i^2.$$
Without this observation the estimate would not follow.
The final assembly uses the fact that
$$\sum a_i(a_{i+1}+a_{i+2}) = \sum a_i a_{i+1} + \sum a_i a_{i+2}.$$
Each summand is bounded by $\sum a_i^2$, so the whole denominator is bounded by $2\sum a_i^2$. Replacing a denominator by a larger quantity would reverse the inequality, so the direction must be checked carefully. Since
$$\sum a_i(a_{i+1}+a_{i+2}) \le 2\sum a_i^2,$$
its reciprocal is at least
$$\frac1{2\sum a_i^2},$$
which gives the required lower bound.
Alternative Approaches
A different route begins by proving the stronger inequality
$$\sum_{i=1}^{n}\frac{a_i}{a_{i+1}+a_{i+2}} \ge \frac n2.$$
Indeed, Engel's inequality gives
$$\sum_{i=1}^{n}\frac{a_i}{a_{i+1}+a_{i+2}} \ge \frac{(a_1+\cdots+a_n)^2} {\sum a_i(a_{i+1}+a_{i+2})},$$
and the denominator does not exceed
$$2\sum a_i^2\le \frac{2}{n}(a_1+\cdots+a_n)^2.$$
This yields the bound $n/2$. Combining it with
$$(\sum a_i)^2\le n\sum a_i^2$$
gives
$$\frac{(\sum a_i)^2}{2\sum a_i^2}\le \frac n2,$$
and the desired inequality follows.
The main proof is preferable because it directly compares the right-hand side with the exact left-hand side. No intermediate estimate by $n/2$ is needed, and every inequality contributes immediately to the target expression.