Knowlpedia

Weighted arithmetic–geometric mean inequality

For a,b≥0 and θ∈(0,1): a^θ b^(1−θ) ≤ θa+(1−θ)b

Proposition (Weighted AM–GM). For all a,b≥0a,b\ge 0 and all θ∈(0,1)\theta\in(0,1),

aθb1−θ≤θa+(1−θ)b.a^\theta b^{1-\theta}\le \theta a+(1-\theta)b.

Proof sketch. Assume a,b>0a,b>0. of −ln⁡-\ln implies

−ln⁡(θa+(1−θ)b)≤−θln⁡(a)−(1−θ)ln⁡(b).-\ln(\theta a+(1-\theta)b)\le -\theta\ln(a)-(1-\theta)\ln(b).

Exponentiating yields θa+(1−θ)b≥aθb1−θ\theta a+(1-\theta)b\ge a^\theta b^{1-\theta}. The cases a=0a=0 or b=0b=0 follow by continuity or direct inspection.

Examples
  • With θ=12\theta=\tfrac12, this becomes ab≤a+b2\sqrt{ab}\le \tfrac{a+b}{2}.
Remarks

Context. This inequality can be derived from convexity of the function x↦−ln⁡xx\mapsto -\ln x on (0,∞)(0,\infty) and is used to prove Hölder-type inequalities.