Vtyjkyuk

Can we prove this inequality $1+sqrt2+cdots+sqrtn<dfrac4n+36sqrtn$ using integrals by parts?

I need to prove this inequality, and I tried this way:

$$beginsplit
int_0^nsqrtx ,dx &=sumlimits_k=1^n int_k-1^k sqrtx, dx\
&=sumlimits_k=1^n int_0^1 sqrtt+k-1 , dt\
&=sumlimits_k=1^n sqrtk-sumlimits_k=1^n int_0^1dfract2sqrtt+k-1,dt
endsplit$$

Thus
$$sumlimits_k=1^n sqrtk=int_0^n sqrtx, dx+sumlimits_k=1^n int_0^1 dfract2sqrtt+k-1, dt$$.

But maybe I didn't find some inequality trick, I cannot find a proper approxmation of the sum of the integrals.

First, see that we have

$$
int_0^n sqrtx ;dx = frac4n6sqrtn
$$

which gives us a "part" of the demanded result. Moreover, it suggests that we should try to prove the following:

$$
sum_k=1^n int_0^1 fract2sqrtt+k-1 ;dt < frac36sqrtn = frac12sqrtn
$$

Evaluating the integral we get:

beginalign
int_0^1 fract2sqrtt+k-1 ;dt &=
int_k-1^k fracx-k+12sqrtx ;dx =
frac12 int_k-1^k sqrtx ;dx - (k-1)
int_k-1^k frac12sqrtx ;dx \
&=
frac13 left(-2ksqrtk + 2ksqrtk-1 + 3sqrtk - 2sqrtk-1right)
endalign

However, for all $k geq 1$, you can prove that:

$$
frac13 left(-2ksqrtk + 2ksqrtk-1 + 3sqrtk - 2sqrtk-1right) <
frac12 left(sqrtk-sqrtk-1right)
$$

So we end up with a telescoping sum and the result we wanted:

$$
sum_k=1^n int_0^1 fract2sqrtt+k-1 ;dt <
frac12 sum_k=1^n left(sqrtk-sqrtk-1right) =
frac12 sqrtn
$$

Hint: Prove that $$int_k-1^ksqrtxdx>dfracsqrtk+sqrtk-12$$ and then sum up to get your desired conclusion. You can just do this manually or generally this follows from a very well-known fact that the trapeziodal rule underestimates integral if the function is concave down. ($f(x) = sqrtx$) is concave down)