\(a^2+b^2+c^2+d^2=\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)+\left(c^2-c+\frac{1}{4}\right)+\left(d^2-d+\frac{1}{4}\right)+\left(a+b+c+d\right)-1\)
\(=\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2+\left(c-\frac{1}{2}\right)^2+\left(d-\frac{1}{2}\right)^2+1\ge1\)
Dấu "=" xảy ra khi và chỉ khi \(a=b=c=d=\frac{1}{2}\)