Question

Chứng minh rằng với mọi a, b, c, d thì: a² + b² + c² + d² +1 \(\ge\) a + b + c + d

Answer 1

\(a^2+b^2+c^2+d^2+1\ge a+b+c+d\)

\(\Leftrightarrow a^2+b^2+c^2+d^2+1-a-b-c-d\ge0\)

\(\Leftrightarrow\left(a^2-a+\dfrac{1}{4}\right)+\left(b^2-b+\dfrac{1}{4}\right)+\left(c^2-c+\dfrac{1}{4}\right)+\left(d^2-d+\dfrac{1}{4}\right)\ge0\)

\(\Leftrightarrow\left(a-\dfrac{1}{2}\right)^2+\left(b-\dfrac{1}{2}\right)^2+\left(c-\dfrac{1}{2}\right)^2+\left(d-\dfrac{1}{2}\right)^2\ge0\)(luôn đúng)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)