MathLabs

Problem 6

Let MM, KK, and LL be points chosen in the interiors of sides ABAB, BCBC, and CACA, respectively, of triangle ABCABC. Prove that the area of at least one of the three triangles △AML\triangle AML, △KBM\triangle KBM, and △LCK\triangle LCK is less than or equal to one-fourth of the area of triangle ABCABC.
Step 5 of 7: Multiply the three strict inequalities
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In plain words

Multiplication turns the cyclic products from the three area formulas into three independent copies of the same one-variable expression t(1−t)t(1-t).

p(1−p)q(1−q)r(1−r)>164p(1-p)q(1-q)r(1-r)>\frac{1}{64}
Detailed analysis

Multiplying the inequalities in step 4 gives p(1−r)q(1−p)r(1−q)>(14)3=164p(1-r)q(1-p)r(1-q)>\left(\frac14\right)^3=\frac1{64}. Rearranging the positive factors yields p(1−p)q(1−q)r(1−r)>164p(1-p)q(1-q)r(1-r)>\frac1{64}.