Inequality with vectors Suppose that we are given a vector $v = a_1\mu_1+a_2\mu_2+...a_n\mu_n$
where $\mu_i$ are linearly independent, but not orthogonal vectors. All $\mu_i$ are of equal Euclidean length.
$\sum_{i=1}^na_i = 1$
Suppose that $\mu_{j_1}, \mu_{j_2},... \mu_{j_k}$ are  $k$ vectors from the original set such that the length of projection of vector $v$ on this $k$ vectors is highest possible.
In other words, you need to choose $k$ vectors from $\mu_1,\mu_2,...,\mu_n$ such that when you project $v$ onto these vectors length of the projection is highest possible. $\mu_{j_1}, \mu_{j_2},... \mu_{j_k}$ are these $k$ vectors.
Lets denote by $v_{\text{proj}}$ projection of vector $v$ onto these $k$ vectors.
I want to prove that $ \underset{a_1,a_2,...a_n}{\arg \min}||v_{\text{proj}}|| = (a_1 = a_2 = ... = a_n  = \frac{1}{n})$
Notice that when we change $a_i$ the choice of the best $k$ vectors also changes. $\sum_{i=1}^na_i = 1$
It is easy to prove if vectors $\mu_i$ are orthogonal, but how do we prove it in general? I came up myself with this problem, so if you show  a counterexample to my claim I will be very grateful.
 A: I formalize your question as follows. Let $X=\{x_1,\dots,x_n\}$ be a linearly independent set of vectors of unit length of $\Bbb R^m$ (necessarily $m\ge n$) and $k\le n$ be a natural number. Put $[n]^k=\{ J\subset  \{1,\dots,n\}:|J|=k \}$ and $[n]=[n]^1$. 
Clearly, $|[n]^k|={n\choose k}$. For any sequence $(a_1,\dots, a_n)$ of non-negative real numbers such that $\sum a_i=1$ let 
$a=\|(a_1,\dots,a_n)\|_2$. By the inequality between quadratic and arithmetic means, $a^2\ge\frac{\left(\sum_{i=1}^n a_i\right)^2}{n},$ and the equality holds iff $a_1=\dots=a_n=\tfrac 1n$. Put $$f(a_1,\dots,a_n)=\max\left\{\left\|\sum_{j\in J}: a_jx_j\right\|_2: J\in [n]^k\right\}.$$ 
The question is whether $f(a_1,\dots,a_n)\ge f\left(\tfrac 1n,\dots, \tfrac 1n\right)$. This is not always true. For instance, let $n=m=3$, $0<\alpha<\frac{\pi}2$ be a small number, $x_1=(1,0,0)$, $x_2=(-\cos\alpha,\sin\alpha,0)$, $x_3=(0,0,1)$, and $k=2$. 
Then $$f\left(\tfrac 13, \tfrac 13, \tfrac 13\right)=\tfrac 13\|x_1+x_3\|_2=\tfrac{\sqrt{2}}{2}> \tfrac 12=\tfrac 12\|x_1\|_2=
f\left(\tfrac 12, \tfrac 12,0\right).$$ 
On the other hand, it is easy to see that $f(\tfrac 1n ,\dots, \tfrac 1n)\le \tfrac kn$ and by a simple averaging argument below we can show that if $n\ge 2$ then $f(a_1,\dots,a_n)\ge \sqrt{\frac {k((k-1)d^2+ (n-k)a^2)}{n(n-1)} },$ where $d$ is the distance from the origin $0$ of $\Bbb R^m$ to the convex hull $\operatorname{conv} X$ of the set $X$. Clearly, $d\ge h$, where $h$ is the distance from the origin to the affine hull $\operatorname{aff} X$ of the set $X$. Since $X$ is linearly independent, $h>0$. More precisely, $h=\tfrac 1{n}\tfrac VS$, where $V$ is content (a counterpart of the volume) of the simplex with vertices $0, x_1,\dots, x_n$ and ans $S$ is content of the simplex with vertices $x_1,\dots, x_n$. 
Given $(a_1,\dots,a_n)$, we can prove the claimed inequality as follows. 
$$\sum_ {J\in [n]^k} \left\|\sum_{j\in J} a_jx_j\right\|_2^2=$$
$$\sum_ {J\in [n]^k} \left(\sum_{j\in J} a_jx_j, \sum_{j\in J} a_jx_j\right)=$$
$$\sum_ {J\in [n]^k} \sum_{i,j\in J} (a_ix_i,a_jx_j)= $$
$$\sum_{i,j\in [n]} \sum_ {J\in [n]^k\,, i,j\in J}  (a_ix_i,a_jx_j)= $$
$$\sum_{i,j\in [n]\,, i\ne j} \sum_ {J\in [n]^k\,, i,j\in J}  (a_ix_i,a_jx_j)+\sum_{i\in [n]^1} \sum_ {J\in [n]^k\,, i\in J}  (a_ix_i,a_ix_i)= $$
$$\sum_{i,j\in [n]\,, i\ne j} {n-2\choose k-2}  (a_ix_i,a_jx_j)+\sum_{i\in [n]} {n-1\choose k-1}  (a_ix_i,a_ix_i)= $$
$${n-2\choose k-2}\sum_{i,j\in [n]}  (a_ix_i,a_jx_j)+ \left({n-1\choose k-1}-{n-2\choose k-2}\right)\left(\sum_{i\in [n]}   (a_ix_i,a_ix_i)\right)=$$
$${n-2\choose k-2}\left(\sum_{i\in [n]} a_ix_i, \sum_{j\in [n]} a_jx_j\right)+\left({n-1\choose k-1}-{n-2\choose k-2}\right)\left(\sum_{i\in [n]} a_i^2(x_i,x_i)\right)\ge $$
$${n-2\choose k-2}d^2+\left({n-1\choose k-1}-{n-2\choose k-2}\right)\left(\sum_{i\in [n]} a_i^2\right)=$$
$${n-2\choose k-2}d^2+{n-2\choose k-1}a^2.$$
Therefore there exists $J\in [n]^k$ such that 
$$f(a_1,\dots,a_n)\ge \left\|\sum_{j\in J} a_jx_j\right\|_2 \ge
\sqrt{\frac{{n-2\choose k-2}d^2+{n-2\choose k-1}a^2}{ {n\choose k}}}
=\sqrt{\frac {k((k-1)d^2+ (n-k)a^2)}{n(n-1)} }.$$
(in the last equality we assumed that $n\ge 2$).
