Question

If a² + b^2= 1, and x^2 + y^2 = 1, show that ax + by <1.

Answer 1

Answer:

a^2+b^2 = 1

x^2+y^2 = 1

multiplying we get

a^2x^2 + b^2y^2 + a^2y^2 + b^2x^2 = 1

(ax + by)^2 -2axby + (ay)^2 + (bx)^2 = 1

(ax+by)^2 + (ay)^2 - 2aybx + (by)^2 = 1

(ax+by)^2 + (ay -bx)^2 = 1

both the terms on LHS are square, hence both the terms are positive

now two possibilities arise

either ay - bx = 0 and ax + by = 1 or vice versa

or ax+by <1 and ay-bx <1 such that sum of their square is 1

in both the cases

ax+by <=1

{note: in question, the inequality to prove is ax+by<1, whereas my solution is ax+by<=1, since no other information/condition is given, I can safely assign following values to a,b,x and y

a= x= 1 and b=y=0

in this case

a^2+b^2=1, x^2+y^2=1, and ax+by =1 which will contradict the original inequality. kindly check it.)