Problem 1

Let

P

be a point inside a given triangle

ABC

, and let

D

E

F

be the feet of the perpendiculars from

P

to the lines

BC

CA

AB

respectively. Find all positions of

P

for which

\dfrac{BC}{PD} + \dfrac{CA}{PE} + \dfrac{AB}{PF}

is least.

Step 3 of 5: Apply the Cauchy–Schwarz inequality

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In plain words

Cauchy–Schwarz trades a product of two variable expressions for a fixed lower bound built only from the (constant) perimeter — a classic way to turn a sum-of-ratios minimization into an equality condition.

\Big(BC\cdot PD + CA\cdot PE + AB\cdot PF\Big)\left(\frac{BC}{PD}+\frac{CA}{PE}+\frac{AB}{PF}\right) \ge (BC+CA+AB)^2

Detailed analysis

Write $u_1=\sqrt{BC\cdot PD}$ , $u_2=\sqrt{CA\cdot PE}$ , $u_3=\sqrt{AB\cdot PF}$ and $v_1=\sqrt{BC/PD}$ , etc.; Cauchy–Schwarz gives $(\sum u_iv_i)^2 \le (\sum u_i^2)(\sum v_i^2)$ , which after simplifying $u_iv_i$ is exactly the displayed inequality.