Problem 4

Let

ABCD

be a convex quadrilateral with

AB=AD+BC

. Suppose that an interior point

P

has distance

h

from the line

CD

and satisfies

AP=h+AD

and

BP=h+BC

. Prove that

\frac1{\sqrt h}\ge\frac1{\sqrt{AD}}+\frac1{\sqrt{BC}}

Step 5 of 5: Return from equality to the inequality

Previous step Next step

In plain words

Any non-extremal circle is smaller; a smaller height makes its reciprocal square root larger.

h\le h_{\max}\ \Longrightarrow\ \frac1{\sqrt h}\ge\frac1{\sqrt{AD}}+\frac1{\sqrt{BC}}

Detailed analysis

The extremal configuration maximizes $h$ for the fixed lengths $AD,BC$ ; every admissible quadrilateral has $h\le h_{\max}$ . Since $x\mapsto1/\sqrt{x}$ decreases for positive x, the equality from Step 4 at $h_{\max}$ becomes $\frac1{\sqrt h}\ge\frac1{\sqrt{AD}}+\frac1{\sqrt{BC}}$ , as required.