Any solution

The question is implied by ISL 2003 A6, and can be done in a similar way. First we normalize the variables. Take $A=\max \{a_i\}$ and $B=\max\{b_i\}$, and there after we transform the variables as follows: $a_i\rightarrow \frac{a_i}{A}, b_i\rightarrow \frac{b_i}{B},c_i\rightarrow\frac{c_i}{\sqrt{AB}}$ (and if $A$ or $B$ is zero, the result is trivial). Note that the conditions and the problem statement are preserved, but now we can assume WLOG that $A=B=1$. So we go ahead and AM-GM the LHS: it suffices to show that $$\sum^n_{i=0} (a_i+b_i) \le 2\sum_{k=0}^{2n}c_k$$. But we note that for each real $r$, the number of terms $\ge r$ on the RHS is always at least that of the LHS. This is not too hard to verify as whenever $a_i,b_j\ge r$, then $c_{i+j}\ge r$. Hence we are done.

Use induction. Worked for me. If can't get I'll post the solution

Can you please send the solution?(with induction)

Let $a_p=max\{(a_0,a_1,\cdots,a_n)\}$ and $b_q=max\{(b_0,b_1,\cdots,b_n)\}$. Then \[\begin{aligned} \left(\sum_{k=0}^{2 n} c_{k}\right)^{2} & \geqslant\left(\sum_{k=q}^{q+n} c_{k}\right)\left(\sum_{k=p}^{p+n} c_{k}\right)=\left(\sum_{i=0}^{n} c_{i+q}\right)\left(\sum_{j=0}^{n} c_{j+p}\right)\\ &\geqslant\left(\sum_{i=0}^{n} \sqrt{a_{i} b_{q}}\right)\left(\sum_{j=0}^{n} \sqrt{a_{p} b_{j}}\right)=\sqrt{a_{p} b_{q}}\left(\sum_{i=0}^{n} \sqrt{a_{i}}\right)\left(\sum_{j=0}^{n} \sqrt{b_{j}}\right) \\ &=\left(\sum_{i=0}^{n} \sqrt{a_{i} a_{p}}\right)\left(\sum_{j=0}^{n} \sqrt{b_{q} b_{j}}\right) \geqslant \sum_{i=0}^{n} a_{i} \cdot \sum_{j=0}^{n} b_{j}. \end{aligned}\]