Prove that $a_1^2+a_2^2+\dots+a_n^2\ge\frac14\left(1+\frac12+\dots+\frac1n\right)$ 
Let $a_1,a_2,\dots,a_n$ be non-negative real numbers such that for any $k\in\mathbb{N},k\le n$
  $$a_1+a_2+\dots+a_k\ge\sqrt k$$
  Prove that
  $$a_1^2+a_2^2+\dots+a_n^2\ge\frac14\left(1+\frac12+\dots+\frac1n\right)$$

I just want to verify my proof. This is my solution:
By definition I have
$$a_1+a_2+\dots+a_{k-1}\ge\sqrt{k-1}$$
So, minimal value of $a_1+a_2+\dots+a_{k-1}$ is $\sqrt{k-1}$. Now I have
$$a_1+a_2+\dots+a_{k-1}+a_k\ge\sqrt k$$
which means that $a_k\ge\sqrt{k}-\sqrt{k-1}$. This is true for any $k\le n$, so for $k=n$ I need to prove that $a_1^2+a_2^2+\dots+a_n^2\ge\frac14\left(1+\frac12+\dots+\frac1n\right)$. Suppose that
$$a_k^2-\frac1{4k}<0$$
If this inequality isn't true for minimal value of $a_k$, then it cannot be true for any other value of $a_k$, so let's check it:
$$
\left( \sqrt k-\sqrt{k-1} \right)^2-\frac1{4k}<0\\
4k\left(k-2\sqrt{k^2-k}+k-1\right)-1<0\\
8k^2-8k\sqrt{k^2-k}-4k-1<0\\
8k^2-4k-1<8k\sqrt{k^2-k}
$$
We can see that RHS if positive for all $k$. Now I need to check on which interval is LHS positive. After solving quadratic inequality I got $8k^2-4k-1>0$ for $k\in\left(-\infty,\frac{1-\sqrt3}{4}\right)\cup\left(\frac{1+\sqrt3}{4},\infty\right)$ which is true for any $k$, so I can square both sides:
$$
64k^4+16k^2+1-64k^3-16k^2+8k<64k^3-64k^2\\
8k+1<0
$$
Which isn't true, so it is contradiction. Thus the proof is completed. Does my proof have mistakes? Did I assumed somethig that may not be true? Is there an easier way?
 A: The argument is right in the end, but the reason why the value $a_k=\sqrt{k}-\sqrt{k-1}$ is minimal is not complete. What we have here is the convex polyhedron $P$ in $\mathbb R^n$ with inequalities
$$
x_1\ge1,\,x_1+x_2\ge\sqrt{2},\,x_1+x_2+x_3\ge\sqrt{3},\dots,\,x_1+\dots+x_n\ge\sqrt{n},
$$
and the exterior of the open ball $B$ given by
$$
x_1^2+\cdots+x_n<r^2=\frac{1}{4}(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}).
$$
The inequality to prove means exactly that the open ball $B$ doesn't meet the convex polyhedron $P$. Thus one must find the point in the boundary of $P$ closest to the origin, and check that point is off the ball $B$. (If I've checked well) that closest point happens to be your point $a=(a_1,a_2,a_3,\dots,a_n)$:
$$
a_1=1,\,a_2=\sqrt{2}-1,\,a_3=\sqrt{3}-\sqrt{2},\dots,\,a_n=\sqrt{n}-\sqrt{n-1}.
$$
And then you can discuss the inequality $(\sqrt{k}-\sqrt{k-1})^2>\frac{1}{4k}$.
The determination of that closest point $a$ is a max conditioned problem (Analysis, Lagrange multipliers...), in this case a bit cumbersome and I hope I haven't made a mistake. I suggest drawing at least the case $n=3$. In any case, the important thing here is the procedure: given any other bound of the same type with a different $r^2$ this procedure always decides if the inequality holds.   
A: let $b(k)=\sqrt{k}-\sqrt{k-1} ,k=1$ to  $n \implies \sum_{i=1 }^k b(i)=\sqrt{k} $
$b(k)=\dfrac{1}{\sqrt{k}+\sqrt{k-1} }>\dfrac{1}{\sqrt{k}+\sqrt{k} }=\dfrac{1}{2\sqrt{k}} $
it is trivial $b(k)$ is mono decreasing function $\implies b(k-1)>b(k)$, ie,
$b(1 )>b(2)> b(3)>...>b(n)>0$
$a_k=b(k)+d_k,\sum a_i=\sum b(i)+\sum d_i \implies \sum d_i \ge 0 $ ie,
$d_1\ge0\\d_1+d_2 \ge 0 \\d_1+d_2 +d_3\ge 0 \\...\\ d_1+d_2+...d_n \ge 0\\$
$b(1)d_1 \ge b(2)d_1 \implies b(1)d_1+b(2)d_2\ge b(2)d_1+b(2)d_2 =b(2)(d_1+d_2) \ge 0 ...\implies \sum_{i=1}^n b(i)d_i \ge b(n)\sum_{i=1}^n d_i \ge 0$
$\sum_{i=1}^n a_i^2 -\sum_{i=1}^n b(i)^2=2\sum_{i=1}^nb(i)d_i+\sum_{i=1}^n d_i^2 \ge 0$
$\sum_{i=1}^n b(i)^2 > \sum_{i=1}^n \left(\dfrac{1}{2\sqrt{i}}\right)^2 = \sum_{i=1}^n \left(\dfrac{1}{4i}\right) \implies \sum_{i=1}^n a_i^2 >\sum_{i=1}^n \left(\dfrac{1}{4i}\right) $
QED
