Problem
ALG-B2-M11-P011 Tangent estimate
#11
★★★★★ Level 5 of 5
Let \(a,b,c\ge0\), \(a+b+c=3\). Prove \[\frac1{1+a^2}+\frac1{1+b^2}+\frac1{1+c^2}\ge\frac32.\]
Hint. Prove \(\frac1{1+x^2}\ge\frac{2-x}{2}\) for \(x\ge0\).
For \(x\ge0\), \[\frac1{1+x^2}-\frac{2-x}{2}=\frac{x(x-1)^2}{2(1+x^2)}\ge0.\] Summing for \(x=a,b,c\), we get \[\sum\frac1{1+a^2}\ge\frac{6-(a+b+c)}2=\frac32.\]
Teaching goal: the student should first recognise the type of estimate, then choose the tool.