\definecolor{A}{RGB}{250,120,0}\color{A}\fbox{A3.} Assume that a,b,c are positive reals such that a+b+c=3. Prove that \definecolor{A}{RGB}{200,0,200}\color{A} \frac{1}{8a^2-18a+11}+\frac{1}{8b^2-18b+11}+\frac{1}{8c^2-18c+11}\le 3.Proposed by [color=#419DAB]ltf0501.
[color=#3D9186]#1734 quadraticsinequalitiesIMOCTaiwan