Problem
ALG-B2-M01-P003 One fraction
#3
★★☆☆☆ Level 2 of 5
Prove that for \(t>0\), \(\frac{t}{t^2+t+1}\le\frac13\).
Hint 1. The denominator is positive.
Hint 2. Reduce it to \((t-1)^2\).
Since \(t^2+t+1>0\), we may multiply by \(3(t^2+t+1)\). We get \(3t\le t^2+t+1\), i.e. \((t-1)^2\ge0\). Equality occurs at \(t=1\).
Module training problem. Method tags: squares, fractions.