Question

Chứng minh rằng với a, b, c, d tùy ý ta luôn có:

\(a^2+b^2+c^2+d^2\ge\left(a+b\right)\left(c+d\right)\)

Answer 1

\(a^2+b^2+c^2+d^2\ge\left(a+b\right)\left(c+d\right)\)

\(\Leftrightarrow2\left(a^2+b^2+c^2+d^2\right)\ge2\left(ac+ad+bc+bd\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2+2d^2-2ac-2ad-2bc-2bd\ge0\)\(\Leftrightarrow\left(a^2-2ac+c^2\right)+\left(a^2-2ad+d^2\right)+\left(b^2-2bc+c^2\right)+\left(b^2-2bd+d^2\right)\ge0\)\(\Leftrightarrow\left(a-c\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2\ge0\)Luôn đúng với mọi \(a;b;c;d\in Z\)