How can I prove $\sqrt{3}+\sqrt{4}+...+\sqrt{n}<\frac{1}{4}n^2$ for all whole numbers $n \geq 3$? $\sqrt{3}+\sqrt{4}+...+\sqrt{n}<\frac{1}{4}n^2 ; n \geq 3$
Induction first step:
$n=3$
$\sqrt{3}<\frac{9}{4}$ which is true
Induction second step:
$n=k$
$\sqrt{3}+\sqrt{4}+...+\sqrt{k}<\frac{1}{4}k^2$
Induction third step:
$n=k+1$
$\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}<\frac{1}{4}(k+1)^2$
From the assumption in the second step we know that: $\sqrt{3}+\sqrt{4}+...+\sqrt{k}<\frac{1}{4}k^2$
From this we can assume that: $\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}<\frac{1}{4}k^2+\sqrt{k+1}$
So we know that $\frac{1}{4}k^2+\sqrt{k+1}$ is bigger than $\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}$. We can then replace $\sqrt{3}+\sqrt{4}+...+\sqrt{k}+\sqrt{k+1}$ in the statement that we want to prove with $\frac{1}{4}k^2+\sqrt{k+1}$
So:
$$\frac{1}{4}k^2+\sqrt{k+1}<\frac{1}{4}(k+1)^2$$
$$16k+16<4k^2+4k+1$$
$$0<4k^2-12k-15$$
When I solve this I get $\left( \frac{3+2\sqrt{6}}{2};+\infty \right)$ for positive numbers
--Edit--
$3\not\in\left( \frac{3+2\sqrt{6}}{2};+\infty \right)$, but we manually proved above that the statement is true for $3$.
In this case, we proved by induction that this statement is true for $n \geq 4$. We don't need to rewrite our proof but only include a new first induction step.
*Induction first step:
$n=4$
$\sqrt{3}+\sqrt{4}<\frac{16}{4}$ which is true
And because we manually proved for $3$ we can say that this statement is true for all whole numbers $n \geq 3$
Is this correct?
--
Thanks

You have a statement $S(n)$. You attempt to show $S(3)$, and that $S(n)$ implies $S(n+1)$ for all $n \geq 3$. Unfortunately this works only for $n \geq 4$.
You solve the conundrum by showing both $S(3)$ and $S(4)$, and since the induction statement is true for $n \geq 4$, $S(n)$ is also true for all $n \geq 5$.
It happens quite often that the induction statement is only true for $n \geq N$. In that case you prove $S(N)$ by hand and this proves $S(n)$ for all $n \geq N$. Then you try to prove $S(n-1)$, $S(N-2)$ and so on.
But $S(N)$ might also be false. In that case, if you find an $m \geq N$ where $S(m)$ is true, you know $S(n)$ is true for all $n \geq m$, and you try to prove $S(m-1)$ and so on as well. That would happen if you tried to prove $\text{LHS}<\text{RHS}-1000000$. You might not find any such $m$ because the statement is false for all $n$.
