BĐT Bunhiacopxki:
Áp dụng cho 6 số(1,1,1,a,b,c)
\(\left(1^2+1^2+1^2\right).\left(a^2+b^2+c^2\right)\ge\left(1a+1b+1c\right)^2\)
Chứng minh:
\(\left(ax+by\right)^2\le\left(a^2+b^2\right).\left(x^2+y^2\right)\)
\(\Leftrightarrow a^2x^2+2axby+b^2y^2\le a^2x^2+a^2y^2+b^2x^2+b^2y^2\)
\(\Leftrightarrow2axby\le a^2y^2+b^2x^2\)
\(\Leftrightarrow\left(ay-bx\right)^2\ge0\)( đpcm )