Với mọi x;y ta luôn có: \(\left(x-y\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2\ge2xy\)
\(\Leftrightarrow\frac{x^2+2xy+y^2}{4}\ge xy\)
\(\Leftrightarrow\left(\frac{x}{2}+\frac{y}{2}\right)^2\ge xy\)
Áp dụng vào bài toán:
\(\left(\frac{a+b}{2}+\frac{c+d}{2}\right)^2=\left(\frac{a+c}{2}+\frac{b+d}{2}\right)^2\ge\left(a+c\right)\left(b+d\right)\) (đpcm)
Dấu "=" xảy ra khi \(a+c=b+d\)