Example 1. In the figure given below, the diagonals AC and BD of the quadrilateral ABCD are perpendicular. Given that AB = 8, BC = 7 and DA = 4, find CD.
Example 2. If x and y are any two real numbers, show that (x2 + y2)2 ≥ xy(x + y)2.
Solution 1. Let O be the point of intersection of the two diagonals AC and BD. Then the four triangles ∆OAB, ∆OBC, ∆OCD and ∆ODA all have right angles at the vertex O. Applying the Pythagorean theorem, we get
∆OAB : (AB)2 = (OA)2 + (OB)2
∆OBC : (BC)2 = (OB)2 + (OC)2
∆OCD : (CD)2 = (OC)2 + (OD)2
∆ODA : (DA)2 = (OD)2 + (OA)2
It follows that
(AB)2 + (CD)2
= (OA)2 + (OB)2 + (OC)2 + (OD)2
= (OB)2 + (OC)2 + (OA)2 + (OD)2
= (BC)2 + (DA)2.
Since AB = 8, BC = 7 and DA = 4, we conclude that
(CD)2 = (BC)2 + (DA)2 − (AB)2
= 72 + 42 − 82
= 1.
Therefore, CD = 1.
Solution 2. We prove the inequality by showing that the left side minus the right side is always
greater than or equal to zero. We compute
(x2 + y2)2 − xy(x + y)2
= x4 + 2x2y2 + y4 – x3y – 2x2y2 – xy3
= x3(x – y) – y3(x – y)
= (x – y)(x3 – y3).
Since x and y are real numbers, either x = y or x > y or x < y.
If x = y, then (x – y)(x3 – y3) = 0.
If x > y, then x3 > y3. Therefore, (x – y)(x3 – y3) > 0.
If x < y, then x3 < y3. Therefore, (x – y)(x3 – y3) > 0.
Thus, in any case, (x – y)(x3 – y3) ≥ 0.
It follows that, (x2 + y2)2 ≥ xy(x + y)2.