Question 1
Let $a,b,c\ge1$ be real. Prove
$$\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right)\le abc.$$
Proposed by Titu Andreescu (credited in the supplied notes).
$$\min\left(\frac{10a^2-5a+1}{b^2-5b+10},\frac{10b^2-5b+1}{c^2-5c+10},\frac{10c^2-5c+1}{a^2-5a+10}\right)\le abc.$$
Proposed by Titu Andreescu (credited in the supplied notes).