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Question Number 83674 by Power last updated on 05/Mar/20

Commented by Power last updated on 05/Mar/20

$$\mathrm{sir}\mathrm{MJS}$$

Answered by mahdi last updated on 05/Mar/20

$$\mid \mathrm{x}\mid +\mid \mathrm{y}\mid ⩾\mid \mathrm{x}+\mathrm{y}\mid$$$$\mid (\mathrm{x}-1)+(\mathrm{x}+1)\mid +\mid (\mathrm{x}-2)+(\mathrm{x}+2)\mid +...+\mid (\mathrm{x}-500)+(\mathrm{x}+500)\mid$$$$⩽\mid \mathrm{x}-1\mid +\mid \mathrm{x}+1\mid +...+\mid \mathrm{x}+500\mid \Rightarrow$$$$\mid 2\mathrm{x}\mid +...+\mid 2\mathrm{x}\mid ⩽1000\Rightarrow$$$$500\mid 2\mathrm{x}\mid =1000\mid \mathrm{x}\mid ⩽1000\Rightarrow \mid \mathrm{x}\mid ⩽1$$$$\Rightarrow -1⩽\mathrm{x}⩽1$$$$1000\stackrel{?}{=}\mid \mathrm{x}-1\mid +\mid \mathrm{x}+1\mid +...+\mid \mathrm{x}+500\mid \stackrel{?}{=}(1-\mathrm{x})+(\mathrm{x}+1)+...+(500-\mathrm{x})+(\mathrm{x}+500)$$$$1000\ne 2\left(\frac{500\times 501}{2}\right)\Rightarrow \mathrm{without}\mathrm{answer}$$$$$$

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