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Question Number 1268 by 314159 last updated on 18/Jul/15

$$\mathbf{Prove}\mathbf{that}\mathbf{AM}>\mathbf{HM}.$$

Answered by prakash jain last updated on 18/Jul/15

$$\mathrm{AM}=\frac{a+b}{2},\mathrm{HM}=\frac{2ab}{a+b}$$$$\mathrm{AM}-\mathrm{HM}=\frac{a+b}{2}-\frac{2ab}{a+b}=\frac{{(a-b)}^2}{2(a+b)}$$$$\mathrm{If}a,b>0\mathrm{then}\mathrm{AM}⩾\mathrm{HM}$$$$\mathrm{AM}⩾\mathrm{HM}\mathrm{is}\mathrm{only}\mathrm{valid}\mathrm{for}+\mathrm{ve}\mathrm{numbers}.$$$$\mathrm{Try}:a=-4,b=2$$

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