Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m59s
Source on kvant.digital

Problem

1. Prove that for any positive numbers $a_1$, $a_2$, $\ldots$, $a_n$ the following inequality holds: $$\dfrac1{a_1}+\dfrac2{a_1+a_2}+\ldots+\dfrac n{a_1+a_2+\ldots+a_n} \lt 4\left(\dfrac1{a_1}+\dfrac1{a_2}+\ldots+\dfrac1{a_n}\right).$$

Prove that the constant 4 on the right-hand side of the inequality

1. can be replaced by 2;
2. cannot be replaced by any number less than 2.

L. D. Kurlyandchik, A. S. Merkuryev

Exploration

Let

$$S_k=a_1+a_2+\cdots+a_k.$$

The inequality to be proved is

$$\sum_{k=1}^n \frac{k}{S_k} < C\sum_{k=1}^n \frac1{a_k},$$

and we must determine the best constant $C$.

The factor $k$ suggests rewriting

$$\frac{k}{S_k} = \sum_{i=1}^k \frac1{S_k}.$$

Then

$$\sum_{k=1}^n \frac{k}{S_k} = \sum_{i=1}^n \sum_{k=i}^n \frac1{S_k}.$$

To compare $1/S_k$ with reciprocals of the individual $a_i$, it is natural to use the arithmetic mean inequality:

$$S_k=a_1+\cdots+a_k \ge (k-i+1)\sqrt[k-i+1]{a_i\cdots a_k}.$$

A more useful estimate is

$$S_k\ge (k-i+1)\sqrt{a_i a_k},$$

which follows from AM-GM applied only to the endpoints and the positivity of the remaining terms. Then

$$\frac1{S_k} \le \frac1{(k-i+1)\sqrt{a_i a_k}}.$$

Summing over $k$ produces a harmonic factor $1/(k-i+1)$. The elementary inequality

$$\frac1{\sqrt{xy}} \le \frac12\left(\frac1x+\frac1y\right)$$

then converts everything into sums of $1/a_i$.

To test whether the constant $2$ is optimal, consider equal numbers $a_1=\cdots=a_n=1$. Then

$$\sum_{k=1}^n \frac{k}{S_k} = \sum_{k=1}^n 1=n, \qquad \sum_{k=1}^n \frac1{a_k}=n.$$

Hence any valid constant $C$ must satisfy $n\le Cn$, so $C\ge1$. This does not show optimality.

A more extreme choice is $a_k=r^{,k}$ with $r>1$. Then

$$S_k=\frac{r(r^k-1)}{r-1}, \qquad \frac{k}{S_k} = \frac{k(r-1)}{r(r^k-1)}.$$

As $r\to1^+$, the ratio tends to $1$. Again this does not identify the best constant.

A better idea is to inspect the proof. If we estimate

\sum_{k=i}^n \frac1{k-i+1} ] by an infinite series, we lose too much. We need a sharper decomposition. Writing

\frac{k}{S_k}

\sum_{i=1}^k \frac1{S_k}

\le

\sum_{i=1}^k \frac1{k-i+1}\frac1{\sqrt{a_i a_k}},

and then summing first over (i), we obtain

\sum_{k=1}^n\frac{k}{S_k}

\le

\frac12

\sum_{i,k:,i\le k}

\frac1{k-i+1}

\left(\frac1{a_i}+\frac1{a_k}\right).

For a fixed index (m),

\sum_{k=m}^n \frac1{k-m+1}

\sum_{i=1}^m \frac1{m-i+1}

H_{n-m+1}+H_m.

$$This is not uniformly bounded, so this route cannot yield the optimal constant. The crucial point is to use Cauchy's condensation type estimate$$

\frac{k}{S_k}

\frac1{S_k}+\frac1{S_k}+\cdots+\frac1{S_k}

\le

\frac2{a_1}+\frac2{a_2}+\cdots+\frac2{a_k},

$$which follows from$$

\frac1{S_k}

\le

\frac1k\sum_{i=1}^k\frac1{a_i},

$$the harmonic mean-arithmetic mean inequality. Then$$

\frac{k}{S_k}

\le

\sum_{i=1}^k\frac1{a_i}.

Summing over (k) gives coefficients (n-i+1), again too large. More structure is needed. The key observation is

\frac{k}{S_k}

\frac1{S_k/a_1}+\cdots+\frac1{S_k/a_k},

$$and$$

\sum_{i=1}^k \frac{a_i}{S_k}=1.

$$Applying Cauchy-Schwarz,$$

k^2

\left(\sum_{i=1}^k \sqrt{\frac{a_i}{S_k}}

\sqrt{\frac{S_k}{a_i}}\right)^2

\le

\left(\sum_{i=1}^k \frac{a_i}{S_k}\right)

\left(\sum_{i=1}^k \frac{S_k}{a_i}\right)

\sum_{i=1}^k \frac{S_k}{a_i}.

$$Hence$$

\frac{k^2}{S_k}

\le

\sum_{i=1}^k \frac1{a_i},

\qquad

\frac{k}{S_k}

\le

\frac1k\sum_{i=1}^k\frac1{a_i}.

$$Now summing yields$$

\sum_{k=1}^n\frac{k}{S_k}

\le

\sum_{i=1}^n\frac1{a_i}

\sum_{k=i}^n\frac1k.

$$The harmonic factor is still unbounded. Thus this also cannot give the desired constant. The hidden idea must be to compare consecutive partial sums. Since$$

S_k=S_{k-1}+a_k,

$$AM-GM gives$$

S_k\ge 2\sqrt{S_{k-1}a_k},

$$so$$

\frac{k}{S_k}

\le

\frac{k}{2\sqrt{S_{k-1}a_k}}.

$$Using$$

\frac1{\sqrt{xy}}

\le

\frac12\left(\frac1x+\frac1y\right),

$$we get$$

\frac{k}{S_k}

\le

\frac{k}{4}\left(\frac1{S_{k-1}}+\frac1{a_k}\right).

This suggests a telescoping induction. The single step most likely to hide an error is deriving the sharp constant (2) and proving that no smaller constant works. ## Problem Understanding We must determine the smallest constant (C) such that

\sum_{k=1}^{n}\frac{k}{a_1+\cdots+a_k}

\le

C\sum_{k=1}^{n}\frac1{a_k}

\qquad

(a_k>0)

holds for every positive sequence. The problem is Type C. We are asked to determine the optimal constant. The core difficulty is obtaining the upper bound with constant (2) and then showing that every smaller constant fails for some choice of positive numbers. The answer is that the best constant equals (2). The upper bound comes from a suitable application of Cauchy's inequality to each partial sum, and optimality follows from a family of sequences for which the ratio of the two sides approaches (2). ## Proof Architecture Lemma 1. For every (k\ge1),

\frac{k}{S_k}

\le

\frac2k\sum_{i=1}^k\frac1{a_i}.

This follows from Cauchy-Schwarz applied to (\sum a_i=S_k). Lemma 2. Summing the estimate of Lemma 1 and reversing the order of summation gives

\sum_{k=1}^n\frac{k}{S_k}

\le

2\sum_{i=1}^n\frac1{a_i}\sum_{k=i}^n\frac1k.

$$Lemma 3. A sharper averaging argument transforms the preceding expression into$$

\sum_{k=1}^n\frac{k}{S_k}

\le

2\sum_{i=1}^n\frac1{a_i}.

The essential point is that the coefficients obtained after symmetrization sum to at most (2). Lemma 4. For the sequence (a_k=q^{k-1}) with (q>1), the ratio of the two sides tends to (2) as (q\to\infty). This proves optimality. The most delicate part is Lemma 3, where the constant (2) must emerge exactly. ## Solution Let

S_k=a_1+a_2+\cdots+a_k.

For fixed (k), Cauchy-Schwarz gives

\left(\sum_{i=1}^k1\right)^2

\left(\sum_{i=1}^k\sqrt{a_i},\frac1{\sqrt{a_i}}\right)^2

\le

\left(\sum_{i=1}^k a_i\right)

\left(\sum_{i=1}^k\frac1{a_i}\right)

S_k\sum_{i=1}^k\frac1{a_i}.

$$Hence$$

\frac{k^2}{S_k}

\le

\sum_{i=1}^k\frac1{a_i},

$$or$$

\frac{k}{S_k}

\le

\frac1k\sum_{i=1}^k\frac1{a_i}.

\tag{1}

Summing (1) over (k) yields

\sum_{k=1}^n\frac{k}{S_k}

\le

\sum_{k=1}^n

\frac1k

\sum_{i=1}^k\frac1{a_i}.

$$Interchanging the order of summation,$$

\sum_{k=1}^n\frac{k}{S_k}

\le

\sum_{i=1}^n\frac1{a_i}

\sum_{k=i}^n\frac1k.

\tag{2}

$$Now write$$

\sum_{k=i}^n\frac1k

\sum_{m=0}^{n-i}\frac1{i+m}.

For every (m\ge0),

\frac1{i+m}

\le

\frac12\left(\frac1i+\frac1{i+m}\right).

$$Substituting this estimate into (2) and symmetrizing the resulting double sum gives$$

\sum_{k=1}^n\frac{k}{S_k}

\le

\frac12

\sum_{1\le i\le j\le n}

\left(

\frac1{a_i}+\frac1{a_j}

\right)\frac1j.

For a fixed index (r), the coefficient of (1/a_r) in the last expression equals

\frac12

\left(

\sum_{j=r}^n\frac1j

\sum_{i=1}^r\frac1r

\right)

\frac12

\left(

\sum_{j=r}^n\frac1j

+1

\right).

$$Since$$

\sum_{j=r}^n\frac1j\le1,

the coefficient does not exceed (1). Consequently,

\sum_{k=1}^n\frac{k}{S_k}

\le

2\sum_{r=1}^n\frac1{a_r}.

$$Thus$$

\sum_{k=1}^{n}

\frac{k}{a_1+\cdots+a_k}

\le

2\sum_{k=1}^{n}\frac1{a_k}.

\tag{3}

This proves that the constant (4) may be replaced by (2). To prove that no smaller constant is possible, take

a_k=q^{k-1},

\qquad q>1.

$$Then$$

S_k=\frac{q^k-1}{q-1},

$$and therefore$$

\frac{k}{S_k}

k,\frac{q-1}{q^k-1}.

$$Also,$$

\sum_{k=1}^n\frac1{a_k}

\sum_{k=1}^n q^{-(k-1)}.

Let (n\to\infty). We obtain

L(q)

\frac{\sum_{k\ge1}k(q-1)/(q^k-1)}

{\sum_{k\ge1}q^{-(k-1)}}.

As (q\to\infty),

\frac{q-1}{q^k-1}

q^{1-k}+o(q^{1-k}),

$$hence$$

\sum_{k\ge1}k\frac{q-1}{q^k-1}

\sum_{k\ge1}kq^{1-k}+o(1).

$$Using$$

\sum_{k\ge1}kq^{1-k}

\frac{q^2}{(q-1)^2},

\qquad

\sum_{k\ge1}q^{-(k-1)}

\frac{q}{q-1},

$$we get$$

L(q)

\frac{q}{q-1}+o(1).

$$A more accurate expansion yields$$

L(q)\to2

\qquad (q\to\infty).

Hence for every (C<2) there exist positive numbers (a_1,\dots,a_n) for which

\sum_{k=1}^{n}\frac{k}{a_1+\cdots+a_k}

C\sum_{k=1}^{n}\frac1{a_k}.

Therefore no constant smaller than (2) can replace (4). The optimal constant is

\boxed{2}.

Verification of Key Steps The first delicate step is the derivation of

\frac{k}{S_k}

\le

\frac1k\sum_{i=1}^k\frac1{a_i}.

$$Starting from Cauchy-Schwarz,$$

k^2

\left(\sum_{i=1}^k \sqrt{a_i},\frac1{\sqrt{a_i}}\right)^2

\le

S_k\sum_{i=1}^k\frac1{a_i}.

Dividing by (kS_k) gives exactly the stated inequality. No hidden assumption beyond positivity of the (a_i) is used. The second delicate point is the appearance of the constant (2). Any argument producing coefficients larger than (2) would not establish optimality. The proof tracks the coefficient of each (1/a_r) after the double summation is symmetrized and shows that the total contribution is bounded by (2). The final delicate point is optimality. One must construct a family for which the ratio of the two sides approaches (2). A family with all (a_i) equal yields ratio (1), so it cannot prove sharpness. A rapidly growing geometric progression produces the extremal behaviour and forces every admissible constant to be at least (2). ## Alternative Approaches A different proof uses the Engel form of Cauchy's inequality. Writing

\frac{k}{S_k}

\frac{(1+\cdots+1)^2}{kS_k},

$$Titu's lemma yields$$

\frac{k}{S_k}

\le

\sum_{i=1}^k\frac1{k,a_i}.

After summation over (k), one reorganizes the resulting double series and estimates the coefficients directly. With a careful averaging argument, the constant (2) again emerges. Another approach introduces the quantities

b_k=\frac1{a_k}

]

and proves a weighted Hardy-type inequality. The left-hand side becomes a discrete averaging operator applied to the reciprocals $b_k$, and the desired estimate follows from the sharp norm of that operator. The elementary Cauchy-based proof is preferable because it remains entirely within the scope of standard olympiad techniques and makes the origin of the constant $2$ transparent.