Odredi sve realne brojeve a1⩾a2⩾⋯⩾a2020⩾0a_1 \geqslant a_2 \geqslant \cdots \geqslant a_{2020} \geqslant 0a1⩾a2⩾⋯⩾a2020⩾0 za koje vrijedi a1+a2+⋯+a2020=1ia12+a22+⋯+a20202=a1.a_1 + a_2 + \cdots + a_{2020} = 1 \quad \text{i} \quad a_1^2 + a_2^2 + \cdots + a_{2020}^2 = a_1.a1+a2+⋯+a2020=1ia12+a22+⋯+a20202=a1.