It is proved that for any a, B, C and D, they belong to R, and there is an inequality (AC + BD) ^ 2

It is proved that for any a, B, C and D, they belong to R, and there is an inequality (AC + BD) ^ 2

In order for the original form to be established, there must be
(ac+bd)^2≤(a^2+b^2)(c^2+d^2)
a^2c^2+b^2d^2+2abcd≤a^2c^2+b^2d^2+a^2d^2+b^2c^2
Must have a ^ 2D ^ 2-2abcd + B ^ 2C ^ 2 ≥ 0
Then (AD BC) ^ 2 ≥ 0
The above formula is tenable, so the original formula is tenable
(a^2+b^2)(c^2+d^2)-(ac+bd)^2
=a^2c^2+a^2d^2+b^2c^2+b^2d^2-a^2c^2-2abcd-b^2d^2
=a^2d^2+b^2c^2-2abcd
=(ad-bc)^2>=0
So (a ^ 2 + B ^ 2) (C ^ 2 + D ^ 2) > = (AC + BD) ^ 2
(ac+bd)^2