Given a, b, c and d are all disticnt and +ve show that if a^2 + b^2 = 1 and x^2 + y^2 = 1 then ax + by < 1?

Given positive and distinct a, b, x, y such that a² + b² = 1 = x² + y² we show that

[0]: ax + by < 1

without relying on triginometry.

Lemma (arithmetic vs. geometric means): Given nonnegative reals u and v we have

[1]: √(uv) ≤ (u+v)/2,

wherein equality holds if and only if u = v. (Proof: Since u and v are reals, 0 ≤ (u–v)² with equality effective exactly when u = v. Adding 4uv to both sides, 4uv ≤ (u + v)². Noting both sides are nonnegative we take square roots then divide by 2 to obtain [1], q.e.d.)

Now let u = a²y² and v = b²x². Then [1] becomes

aybx ≤ (a²y² + b²x²) /2.

Multiplying by 2, then adding a²x² + b²y² and factoring both sides yields

(ax + by)² ≤ (a² + b²)(x² + y²),

where we note that the right-hand side is just 1, as per givens. Taking square roots

[3]: ax + by ≤ 1,

which will be [0] once we eliminate equality. To do so, recall that u = a²y² and v = b²x². So we need to eliminate the possibility that a²y² = b²x², which, after adding a²x² to both sides, simplifies to

[4]: a² = x²

because we were given that a² + b² = 1 = x² + y². But we were also given that a, b, x, y are distinct and positive, so in particular we cannot have [4]. Thus having eliminated equality from [3], our proof is complete.

My proof relies on c^2 + d^2 > 2cd if c and d are distinct

which follows from (c-d)^2 > 0 if c and d are distinct

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

Adding the two we have a^2 + x^2 + b^2 + y^2 = 2

And using the result above we have 2ax + 2by < a^2 + x^2 + b^2 + y^2 = 2

so ax + by < 1

thus all in 1st quadrant : x = cos Θ , y = sin Θ , a = cos µ , b = sin µ...then ax + by = cos Θ cos µ + sin Θ sin µ ≡ cos ( Θ - µ ) < 1 when Θ - µ ╪ 0