Mathematical competition questions are difficult to answer, and partial derivatives should be used normally.

The distances from the three vertices A, B and C of a right triangle surrounded by three straight lines to AB, BC and AC are D 1, D2 and D3, respectively, as follows:

| AB | * d 1+| BC | * D2+| AC | * D3 = 2S(ABC)

And (| AB | * D 1+| BC | * D2+AC * D3) 2.

So d12+d22+d32 > = 4s (ABC)/(| AB | 2+| BC | 2+| AC | 2) =128/5.

The equal sign is true if and only if |AB|/d 1=|BC|/d2=|AC|/d3.

You should be able to handle it yourself, right? Hehe, I wonder if it will help.