Question

Chúng minh rằng:\(a^2\)+\(b^2\)+\(c^2\)+\(d^2\)≥\(\frac{1}{4}\) với a+b+c+d=1

Answer 1

\(a+b+c+d=1\Rightarrow\frac{a}{2}+\frac{b}{2}+\frac{c}{2}+\frac{d}{2}=\frac{1}{2}\)

Ta có:

\(a^2+b^2+c^2+d^2-\frac{1}{2}=a^2+b^2+c^2+d^2-\frac{a}{2}-\frac{b}{2}-\frac{c}{2}-\frac{d}{2}\)

\(=a^2-\frac{a}{2}+\frac{1}{16}+b^2-\frac{b}{2}+\frac{1}{16}+c^2-\frac{c}{2}+\frac{1}{16}+d^2-\frac{d}{2}+\frac{1}{16}-\frac{1}{4}\)

\(=\left(a-\frac{1}{4}\right)^2+\left(b-\frac{1}{4}\right)^2+\left(c-\frac{1}{4}\right)^2+\left(d-\frac{1}{4}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(\Rightarrow a^2+b^2+c^2+d^2-\frac{1}{2}\ge-\frac{1}{4}\)

\(\Rightarrow a^2+b^2+c^2+d^2\ge\frac{1}{4}\)

Dấu "=" xảy ra khi \(a=b=c=d=\frac{1}{4}\)