Problem
ALG-B3-M09-P007 Cannot Be Above Everywhere
#7
★★★☆☆ Level 3 of 5
Let \(f\) be additive and \(f(x)\ge x\) for all \(x\). Prove that \(f(x)=x\).
Apply the condition to \(-x\).
For \(x\), \(f(x)\ge x\). For \(-x\), \(f(-x)\ge -x\), i.e. \(-f(x)\ge -x\), so \(f(x)\le x\). Therefore \(f(x)=x\) for all \(x\).
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