Question 891975: Can, you help me prove this inequality with Cauchy-Schwarz:
Prove that if a,b,c are less than 1 and a+b+c=2 (all variables denote positive real numbers), then
abc/((1-a)(1-b)(1-c))≥8
Thanks
Answer by richard1234(7193)   (Show Source):
You can put this solution on YOUR website! Note that a, b, and c are the sides of a triangle, from the given conditions, and the semiperimeter is 1.
Cross multiplying gives (1-b)(1-c) \le \frac{abc}{8}) . However, disregarding units, the LHS is equal to  , 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  .
Replacing A with rs, the inequality becomes  which is true (equality iff a = b = c = 2/3).
As for a Cauchy-Schwarz inequality solution, I haven't found one yet.
|
|
|
