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 6 of 7: Apply the universal quadratic bound
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In plain words

The parabola t(1−t)t(1-t) peaks at t=12t=\frac12, so no product of the three factors can exceed 164\frac1{64}.

t(1−t)≤14 (0<t<1) ⟹ p(1−p)q(1−q)r(1−r)≤164t(1-t)\le\frac14\ (0<t<1)\ \Longrightarrow\ p(1-p)q(1-q)r(1-r)\le\frac1{64}
Detailed analysis

For 0<t<10<t<1, (t−12)2≥0(t-\frac12)^2\ge0 gives t(1−t)≤14t(1-t)\le\frac14. Applying this separately to t=p,q,rt=p,q,r yields p(1−p)q(1−q)r(1−r)≤(14)3=164p(1-p)q(1-q)r(1-r)\le(\frac14)^3=\frac1{64}, contradicting step 5.