I tried to solve this using the smallest but haven't succeeded. Could you give some hints?

If $x_1\geq x_2\geq \cdots \geq x_n \geq 0$ and $$\frac{x_1}{\sqrt{1}}+\frac{x_2}{\sqrt{2}}+\cdots+\frac{x_n}{\sqrt{n}}=1$$ then prove $$x_1^2+x_2^2+\cdots+x_n^2\leq 1$$

(original problem image)

  • 3
    $\begingroup$ What have you tried? What do you mean by "using the smallest"? Have you found any equality conditions? $\endgroup$ – Calvin Lin May 3 '20 at 5:50

Let's start with:

$$\sum_{i=1}^n \frac{x_i}{\sqrt i}=1\tag{1}$$

...assuming that:

$$x_1\ge x_2\ge \dots\ge x_n\ge 0\tag{2}$$

If you square (1) you get:

$$\sum_{i=1}^n\frac{x_i^2}i+\sum_{1\le i\lt j\le n}\frac{2x_ix_j}{\sqrt i \sqrt j}=1\tag{3}$$

It is obvious that for $i<j$ and (2):

$$\frac{2x_ix_j}{\sqrt i \sqrt j}\ge\frac{x_jx_j}{\sqrt j \sqrt j}=\frac{x_j^2}j\tag{4}$$

By replacing (4) into (3):

$$\sum_{i=1}^n\frac{x_i^2}i+\sum_{1\le i\lt j\le n}\frac{x_j^2}{j}\le 1\tag{5}$$

By expanding (5):

$$\left(x_1^2+\frac{x_2^2}2+\frac{x_3^2}3+\dots +\frac{x_n^2}n \right)+\tag{first item in (5)}$$

$$\left(\frac{x_2^2}2+\frac{x_3^2}3+\dots +\frac{x_n^2}n \right)+\tag{second item, i=1}$$

$$\left(\frac{x_3^2}3+\dots +\frac{x_n^2}n \right)+\tag{second item, i=2}$$


$$+\left(\frac{x_n^2}n \right)\tag{second item, i=n-1}$$

$$\le 1$$

If you simplify the last expression you get:

$$x_1^2+x_2^2+x_3^2+\dots + x_n^2\le 1$$


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