Q:

Comparing two optimization problems

Consider the following two optimization problems with all $a_i$ and $b_i$ non-negative. 
$\max\limits_{x_i \in [0,1]} \sum\limits_{1\leq i\leq n} a_ix_i -
 (\sum\limits_{1\leq i\leq n} b_ix_i)^2 $ and
$\max\limits_{x_i \in [0,1]} \sum\limits_{1\leq i\leq n} a_ix_i -
 \sum\limits_{1\leq i\leq n} b_i^2x_i^2 $. 
Clearly, the second optimization will have a larger value than the first one because at the optimum of the first, the value of the second objective will be greater. 
However, can we also show that the optimum solutions of the second are greater than the first? 
That is, for all $i$, is $x^{1}_i \leq x^{2}_i$ where $x^{1}_i$ and $x^{2}_i$ denotes the optimum $x_i$ for the first and second optimizations respectively? 

A:

Edited answer for the additional assumption $a_i, b_i \geq 0$
A trivial case is when $a_i=b_i=0$ for all $i$, in which case every vector $x \in [0,1]^n$ is optimal for both problems, and we cannot say that a particular solution vector to one bounds the solution vector of the other. 
Claim: If $a_i>0, b_i \geq 0$ for all $i \in \{1, ..., n\}$, then the unique solution $(x_i^{*(2)})_{i=1}^n$ for problem P2 satisfies: 
$$ x_i^{*(2)} =\left\{ \begin{array}{ll}
\min\left[1, \frac{a_i}{2b_i^2}\right] &\mbox{ if $b_i > 0$} \\
1  & \mbox{ if $b_i=0$} 
\end{array}
\right.$$
Further, any (possibly non-unique) solution $(x_i^{*(1)})_{i=1}^n$ to problem P1 must satisfy $x_i^{*(1)}\leq x_i^{*(2)}$ for all $i \in \{1, ..., n\}$. 
Proof: The problem P2 is separable across variables and it is easy to verify its (unique) solution. Now let $(x_i^{*(1)})_{i=1}^n$ be a (possibly non-unique) solution to P1. Fix $i \in \{1, .., n\}$. We want to show $x_i^{*(1)} \leq x_i^{*(2)}$. 
Case 1: Suppose $b_i=0$. Then $x_i^{*(2)}=1$.  But by inspection of problem P1, it is clear we must have $x_i^{*(1)}=1$ (recall that $a_i>0$). So the result trivially holds in this case. 
Case 2: Suppose $b_i> 0$ but $x_i^{*(1)}=0$.  Then the result trivially holds since $0 \leq x_i^{*(2)}$. 
Case 3:  Suppose $b_i > 0$ and $x_i^{*(1)}>0$. Since $x_i^{*(1)}>0$, the optimality condition for probem P1 requires:
$$ \frac{\partial \left[\sum_{j=1}^n a_j x_j - (\sum_{j=1}^n b_jx_j)^2 \right]}{\partial x_i}\left|_{(x_i^{*(1)})}\right. \geq 0 $$
Indeed, if this were not the case, we could decrease $x_i^{*(1)}$ by a small amount and then improve on our "optimal" objective value. It follows that: 
$$ 0 \leq a_i - 2b_i\left(\sum_{j=1}^n b_j x_j^{*(1)}\right) \leq a_i - 2b_i^2x_i^{*(1)}$$
where the final inequality holds because $b_j\geq 0$ and $x_j^{*(1)}\geq 0$ for all $j$. Since $b_i> 0$, we obtain: 
$$ x_i^{*(1)} \leq \frac{a_i}{2b_i^2}$$
But we already know $x_i^{*(1)} \leq 1$, so we obtain 
$$ x_i^{*(1)} \leq \min\left[1, \frac{a_i}{2b_i^2}\right] = x_i^{*(2)}$$
Original answer for the case $a_i, b_i \in \mathbb{R}$
You suggest the second problem will have a larger max value, but this is not always true.  Consider $n=2$ and let $a_1=a_2=1$, $b_1=1, b_2=-1$. 
P1:  $\mbox{Max}_{(x_1,x_2) \in [0,1]^2} [(x_1+x_2) - (x_1-x_2)^2]  \implies x_1^*=x_2^*=1$ and the max value is 2.
P2: $\mbox{Max}_{(x_1,x_2) \in [0,1]^2} [(x_1+x_2) - x_1^2 - x_2^2] \implies x_1^*=x_2^*=1/2$ and the max value is $1/2$.
On the other hand, if we change to $b_1=b_2=1$, then problem P2 will not change, but P1 will change and its new max value is $1/4$, smaller than that of P2.  So we cannot say that one problem will always bound the other. 

