Maths Olympiad Prep

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Geometry Difficulty 7.6 National olympiad, round 2 Prove it Turkey

Let aa, bb, cc be the lengths of the sides of a triangle; rar_a, rbr_b, rcr_c be the corresponding exradii, respectively, and rr be the inradius. Prove that
a+b+c2a2+b2+c2≤ra2+rb2+rc2ra+rb+rc−3r \frac{a+b+c}{2\sqrt{a^2+b^2+c^2}} \leq \frac{\sqrt{r_a^2+r_b^2+r_c^2}}{r_a+r_b+r_c-3r}

Solution

Note that
a+b+c2r=b+c−a2ra=a+b+c2ra−ara=a+b+c2rb−brb=a+b+c2rc−crc. \frac{a+b+c}{2}r = \frac{b+c-a}{2}r_a = \frac{a+b+c}{2}r_a - ar_a = \frac{a+b+c}{2}r_b - br_b = \frac{a+b+c}{2}r_c - cr_c.
Therefore, we have 3r(a+b+c)2=a+b+c2(ra+rb+rc)−ara−brb−crc\frac{3r(a+b+c)}{2} = \frac{a+b+c}{2}(r_a + r_b + r_c) - ar_a - br_b - cr_c and hence
ara+brb+crc=a+b+c2(ra+rb+rc−3r). ar_a + br_b + cr_c = \frac{a+b+c}{2}(r_a + r_b + r_c - 3r).
On the other hand, the Cauchy-Schwarz inequality implies
ara+brb+crc≤a2+b2+c2⋅ra2+rb2+rc2 ar_a + br_b + cr_c \le \sqrt{a^2 + b^2 + c^2} \cdot \sqrt{r_a^2 + r_b^2 + r_c^2}
and the result follows.

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