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Question Number 146730 by puissant last updated on 15/Jul/21

Commented by puissant last updated on 15/Jul/21

$$\mathrm{merci}$$

Commented by Olaf_Thorendsen last updated on 15/Jul/21

$$\mathrm{Inegalite}\mathrm{triangulaire}:$$$$\mid a+b-c-d\mid ⩽\mid a-c\mid +\mid b-d\mid$$$$\Rightarrow \frac{\mid a+b-c-d\mid }{2}⩽\frac{\mid a-c\mid +\mid b-d\mid }{2}$$$$\Rightarrow \frac{a+b+c+d+\mid a+b-c-d\mid }{2}$$$$⩽\frac{a+b+c+d+\mid a-c\mid +\mid b-d\mid }{2}$$$$=\frac{a+c+\mid a-c\mid }{2}+\frac{b+d+\mid b-d\mid }{2}$$$$\max (a+b,c+d)⩽\max (a,c)+\max (b,d)$$

Answered by Olaf_Thorendsen last updated on 15/Jul/21

$$\mathrm{Differentes}\mathrm{approches}\mathrm{mais}\mathrm{une}$$$$\mathrm{approche}\mathrm{elegante}\mathrm{consiste}\mathrm{a}\mathrm{remarquer}$$$$\mathrm{que}\max (x,y)=\frac{x+y+\mid x-y\mid }{2}.$$$$\mathrm{Il}\mathrm{ne}\mathrm{reste}\mathrm{alors}{\mathrm{qu}}^{\prime }\mathrm{a}\mathrm{utiliser}$$$${\mathrm{l}}^{\prime }\mathrm{inegalite}\mathrm{triangulaire}.$$

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