Ineqality problem

Question

My mentor gave this problem to our class. later on he also told the solution to the problem.but i did not understood the solution as it was done in a very complicated way. so can anyone help me the solve this problem with proper explanation

$$ x_1\geqslant x_2\geqslant x_3\geqslant...\geqslant x_n ....$$

$$\text{where } x_i \in \mathbb{R} $$

such that for $n\geqslant1$ ; $$\frac{x_1}{1}+\frac{x_4}{2}+\frac{x_9}{3}+...+\frac{x_{n^2}}{n}\leqslant1$$

Then PROVE that

for $k\geqslant1$; $$\frac{x_1}{1}+\frac{x_2}{2}+\frac{x_3}{3}+...+ \frac{x_k}{k}\leqslant3 $$

As Einstein said, any explanation should be as simple as possible, but not simpler. Maybe your mentor's solution was as simple as possible? — Paul, Apr 05 '17 at 13:37
I agree with @quasi that there is something weird in your expression, what if $x_{n+1}= \ldots = x_{n^2}=n$ ? there is no condition on these terms in your problem. — Surb, Apr 05 '17 at 13:49
If the descending condition continues to $x_{n^2}$, it might make sense. — quasi, Apr 05 '17 at 13:52

Aravind · Accepted Answer · 2017-04-05T15:11:46.070

Let's try to prove by induction that $\sum_{i=1}^{k} \dfrac{x_i}{i} \leq 3-f(k)$ for some suitable function $f(k) \geq 0$.

Let's consider the induction step.

We want to prove: $\dfrac{x_k}{k} \leq f(k-1)-f(k)$ for a suitable function $f$. Thus, we need an upper bound for $x_k/k$. For this, let's use the given information.

We obtain: $x_{n^2}H_n \leq 1$, where $H_n=\sum_{i=1}^{n}\dfrac{1}{i}$.

Consider the function $A_n=\sum_{t=n+1}^{\infty} \dfrac{1}{t^2H_t}$. This is well-defined because the series is easily seen to be convergent. Observe that $A_{n-1}-A_n=\dfrac{1}{n^2H_n}$.

Now the upper bound for $x_k/k$: write $n=\lfloor \sqrt{k} \rfloor$.

Then $\dfrac{x_k}{k} \leq \dfrac{x_{n^2}}{n^2} \leq \dfrac{1}{n^2H_n}$, and we have: $\dfrac{x_k}{k} \leq A_{n-1}-A_n$.

Thus, we can now set $f(k)=A_{\lfloor \sqrt{k} \rfloor}$, and make the induction work.

Now the base case: We want: $x_1 \leq 3-A_1$ and since $x_1 \leq 1$, it is sufficient to show that $A_1 \leq 2$, which is doable, for eg: $A_1 \leq \dfrac{\pi^2}{6}-1 <1$.

score 2 · Answer 2 · answered Apr 05 '17 at 16:20

For a given $k \ge 1$ choose $n \ge 1$ such that $k < (n+1)^2$. Then $$ \frac{x_1}{1}+\frac{x_2}{2}+\frac{x_3}{3}+...+ \frac{x_k}{k} \le \sum_{j=1}^n \sum_{l=j^2}^{(j+1)^2-1} \frac{x_l}{l} \le \sum_{j=1}^n (2j+1) \frac{x_{j^2}}{j^2} \le 3 \sum_{j=1}^n \frac{x_{j^2}}{j} \le 3 \, , $$ using $$ \frac{2j+1}{j^2} \le \frac 3j $$ for all positive integers $j$, which is easy to verify.

Ineqality problem

2 Answers2