Inequality

[brianlee7799]

If a^2+b^2=1 and c^2+d^2=1, prove that ac+bd <= 1.
I tried to find the maximum value for the first two equations that would result in the largest ac+bd, but I would like to know if there is a more formal solution. Thanks for your help!!

 

[Deleted member 4993]

If a^2+b^2=1 and c^2+d^2=1, prove that ac+bd <= 1.
I tried to find the maximum value for the first two equations that would result in the largest ac+bd, but I would like to know if there is a more formal solution. Thanks for your help!!

\(\displaystyle a^2 - 2ac + c^2 + b^2 - 2bd + d^2 = 2 - 2 (ac + bd)\)

and continue ...

 

[brianlee7799]

(a-c)^2+(b-d)^2=2(1-ac-bd)
.5[(a-c)^2+(b-d)^2]=1-ac-bd
.5[(a-c)^2+(b-d)^2]-1=-(ac+bd)
-0.5[(a-c)^2+(b-d)^2]+1=ac+bd
Thus, ac+bd<=1
QED

Correct me if I made a mistake.
Thank you so much Subhotosh Khan!!