\(A=\sqrt{2\left(\sqrt{x^2+y^2}+x\right)\left(\sqrt{x^2+y^2}+y\right)}-\sqrt{x^2+y^2}-x-y+2020\)\(=\sqrt{2\left(x^2+y^2+\left(x+y\right)\sqrt{x^2+y^2}+xy\right)}-\sqrt{x^2+y^2}-x-y+2020\)\(=\sqrt{\left(x^2+y^2+2xy\right)+2\left(x+y\right)\sqrt{x^2+y^2}+x^2+y^2}-\sqrt{x^2+y^2}-x-y+2020\)\(=\sqrt{\left(x+y\right)^2+2\left(x+y\right)\sqrt{x^2+y^2}+\left(x^2+y^2\right)}-\sqrt{x^2+y^2}-x-y+2020\)\(=\sqrt{\left(x+y+\sqrt{x^2+y^2}\right)^2}-\sqrt{x^2+y^2}-x-y+2020\)\(=x+y+\sqrt{x^2+y^2}-\sqrt{x^2+y^2}-x-y+2020=2020\)

\(A=\frac{\left(1+a\right)\left(1+b\right)\left(1+c\right)}{\left(1-a\right)\left(1-b\right)\left(1-c\right)}\).

Ta có:

\(1-a=a+b+c-a\).(vì \(a+b+c=1\)).

\(\Leftrightarrow1-a=b+c\).

Chứng minh tương tự, ta được:

\(1-b=c+a\); \(1-c=a+b\). Do đó:

\(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(b+c\right)\left(c+a\right)\left(a+b\right)\).

Lại có:

\(1+a=a+b+c+a\)(vì \(a+b+c=1\)).

\(\Leftrightarrow1+a=\left(a+b\right)+\left(a+c\right)\).

Chứng minh tương tự, ta được:

\(1+b=\left(a+b\right)+\left(b+c\right)\); \(1+c=\left(a+c\right)+\left(b+c\right)\),.

Do đó \(\left(1+a\right)\left(1+b\right)\left(1+c\right)=\left[\left(a+b\right)+\left(a+c\right)\right]\left[\left(a+b\right)+\left(b+c\right)\right]\left[\left(a+c\right)+\left(b+c\right)\right]\)

Lúc đó: 

\(A=\frac{\left[\left(a+b\right)+\left(a+c\right)\right]\left[\left(a+b\right)+\left(b+c\right)\right]\left[\left(a+c\right)+\left(b+c\right)\right]}{\left(b+c\right)\left(c+a\right)\left(a+b\right)}\).

Đặt \(a+b=x,b+c=y,c+a=z\left(x,y,z>0\right)\) thì \(x+y+z=2\left(a+b+c\right)=2\). Lúc đó:

\(A=\frac{\left(x+z\right)\left(x+y\right)\left(z+y\right)}{yzx}\).

Vì \(x,y>0\)nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:

\(x+z\ge2\sqrt{xz}\left(1\right)\).

Chứng minh tương tự, ta được:

\(x+y\ge2\sqrt{xy}\left(2\right)\);

\(z+y\ge2\sqrt{zy}\left(3\right)\).

Từ (1), (2), (3), ta được:

\(\left(x+z\right)\left(x+y\right)\left(z+y\right)\ge8\sqrt{xy.yz.zx}=8xyz\).

\(\Rightarrow\frac{\left(x+z\right)\left(x+y\right)\left(z+y\right)}{yzx}\ge\frac{8xyz}{xyz}=8\).

\(\Rightarrow A\ge8\).

Dấu bằng xảy ra.

\(\Leftrightarrow x=y=z>0\Leftrightarrow a+b=b+c=c+a>0\Leftrightarrow a=b=c>0\).

Mà \(a+b+c=1\)nên \(a=b=c=\frac{1}{3}\).

Vậy \(minA=8\Leftrightarrow a=b=c=\frac{1}{3}\).

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