Question 891975
Note that a, b, and c are the sides of a triangle, from the given conditions, and the semiperimeter is 1.


Cross multiplying gives *[tex \large (1-a)(1-b)(1-c) \le \frac{abc}{8}]. However, disregarding units, the LHS is equal to *[tex \large A^2], where A is the area of the triangle, and the RHS equals AR/2, using the area formula A = abc/4R, where R is the circumradius. The inequality reduces to proving *[tex \large A \le \frac{R}{2}].


Replacing A with rs, the inequality becomes *[tex \large r \le \frac{R}{2} \Rightarrow 2r \le R] which is true (equality iff a = b = c = 2/3).

As for a Cauchy-Schwarz inequality solution, I haven't found one yet.