Vtyjkyuk

let $x_iin N^+$,and such $1=x_0le x_1le x_2lecdotsle x_n$ show that
$$sum_i=1^ndfracsqrtx_i-x_i-1x_ilesum_i=1^n^2dfrac1i-dfrac12$$

maybe can use C-S,But I can't it

I think the following can help.

Let for $n>1$ we have $x_nleq n^2$.

Thus, $$sum_i=1^nfracsqrtx_i-x_i-1x_ileqsum_i=1^nfracx_i-x_i-1x_i=sum_i=1^nsum_j=1^x_i-x_i-1frac1x_ileq$$
$$leqsum_i=1^nsum_j=1^x_i-x_i-1frac1x_i-1+j=sum_i=x_0+1^x_nfrac1ileqsum_i=2^n^2frac1i<sum_i=1^n^2frac1i-frac12.$$
If first $x_k>n^2$.

Thus, $$sum_i=k^nfracsqrtx_i-x_i-1x_i<sum_i=k^nfrac1sqrtx_i<sum_i=k^nfrac1n<1$$ and
$$sum_i=1^k-1fracsqrtx_i-x_i-1x_i<sum_i=2^n^2frac1i.$$

I am going to treat $x_i$ as a real variable. Then, we can differentiate the terms with respect to $x_i$:

$$fracddx_ifracsqrtx_i-x_i-1x_i$$

It is easy to show that the derivative is $ 0$ when $x_i=2x_i-1$, and it is also easy to show that it is a maximum (take the 2nd derivave, or just look at the cases $x_i=x_i-1$ and $x_irightarrowinfty $ .

So replacing $x_i=2x_i-1$ we get that each term becomes $frac12sqrtx_i-1$.

Note too, that when using this values for $x_i$, the smaller the values we choose, the term becomes bigger, so there is no doubt that the sum is maxed when we use $x_i=2^i$, therefore:

$$sum_i=1^n fracsqrtx_i-x_i-1x_i leqsum_i=0^n frac12sqrt2^i $$

So if show the worst case we are done. This last series can be easily solved as a geometric one, so the problem is now reduced to showing that:

$$(1+frac 1sqrt2)(1-(frac 1sqrt2)^n+1) leq sum_i=1^n^2 frac 1i - frac12 quad mathbf(1) $$

I will do it by induction. The case $n=1$ is easy. Now we take $(1)$ as our hypothesis, and we want to prove that if the case for $n$ is true, then the case for $n+1$ is true. Starting by the $n+1$ case in the right hand side, we have that:

$$sum_i=1^(n+1)^2 frac 1i - frac12 = sum_i=1^n^2 frac 1i - frac12 +sum_i=n^2+1^(n+1)^2 frac 1i $$

Now we use $(1)$ with the sum up to $n^2$ (and the negative one half), and since we can replace any quantity with a smaller one and the inequality will still be valid, I will replace the latter sum by the number of terms of the sum (expand $(n+1)^2$ to see that the number of terms is $2n+1$ ) multiplied by the smallest term, so now we have to show that:

$$ (1+frac 1sqrt2)(1-(frac 1sqrt2)^n+1)+ frac2n+1(n+1)^2geq (1+frac 1sqrt2)(1-(frac 1sqrt2)^n+2) $$

Rearranging:

$$frac2n+1(n+1)^2geq frac 12 (frac1sqrt2)^n+1 quad mathbf(2)$$

And here I will use a second induction. First, you show that the inequality holds for $n in text1,2,3,4,5$, and using $(2)$ as the hypothesis, we work the left hand side for the case $n+1$ as follows:

$$frac2n+3(n+2)^2 gt frac2n+3sqrt2(n+1)^2 $$

This last inequality, holds for $n gt 5$, because for these values of n, we have that $sqrt[4]2(n+1) gt n+2 $ (That's why I trated the other 5 cases separately). So now we can replace the hypothesis to get:

$$frac2n+3sqrt2(n+1)^2 = frac 12sqrt2 (frac 1sqrt2)^n+1 + frac22sqrt2(n+1)^2 gt frac 12 (frac 1sqrt2)^n+2 $$

Therefore,

$$frac2n+3(n+2)^2 gt frac 12 (frac 1sqrt2)^n+2 $$

This means the $n+1$ case is true, then the hypothesis $(2)$ is true for all natural $n$, and the same goes to the hypothesis $(1)$ and the proof is completed:

$$sum_i=1^n fracsqrtx_i-x_i-1x_i leqsum_i=0^n frac12sqrt2^i leq sum_i=1^n^2 frac 1i - frac12 $$