An Area Inequality

Let a, b, u, v be positive real numbers, such that av - bu = 1. Prove that

a² + b² + u² + v² + au + bv ≥ 3/2.

There may be many ways to prove the required inequality, but one starts with a recognition of the expression av - bu as the area of the parallelogram formed by vectors (a, b) and (u, v):

parallelogram formed by two vectors

How does this help? Let A = a² + b² + u² + v² + au + bv. Then

2A = (a + u)² + (b + v)² + a² + b² + u² + v².

Consider one of the triangles in the diagram. Denote its sides X, Y, Z so that, by the Pythagorean theorem

X² = a² + b²,
Y² = u² + v²,
Z² = (a + u)² + (b + v)²,

and 2A = X² + Y² + Z².

Now, we come to a

Lemma

In a triangle with sides X, Y, Z, area S satisfies

S ≤ (X² + Y² + Z²) / 6.

Proof

Let α be the angle formed by X and Y. Then

S = X·Y·sinα/2 = 2X·Y·sinα/4 ≤ [(X - Y)² + 2XY] / 4 = (X² + Y²) / 4.

(This is so simply because, if α is an angle in a triangle then, 0 < sinα ≤ 1, and 0 ≤ (X - Y)².)

Similar inequalities hold for pairs X, Z and Y, Z:

S = ≤ (X² + Y²) / 4,
S = ≤ (Y² + Z²) / 4,
S = ≤ (Z² + X²) / 4.

Adding the three proves the lemma.

Proof of the inequality

Let T be the area of the parallelogram in the diagram. From Lemma,

T/2 ≤ (X² + Y² + Z²) / 6, i.e.
T ≤ (X² + Y² + Z²) / 3.

But, as the problem stipulates, T = 1, so that

X² + Y² + Z² ≥ 3.

But X² + Y² + Z² = 2A = 2(a² + b² + u² + v² + au + bv), implying

a² + b² + u² + v² + au + bv ≥ 3/2.

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