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Question-146730




Question Number 146730 by puissant last updated on 15/Jul/21
Commented by puissant last updated on 15/Jul/21
merci
$$\mathrm{merci} \\ $$
Commented by Olaf_Thorendsen last updated on 15/Jul/21
Inegalite triangulaire :  ∣a+b−c−d∣ ≤ ∣a−c∣+∣b−d∣  ⇒ ((∣a+b−c−d∣)/2) ≤ ((∣a−c∣+∣b−d∣)/2)  ⇒ ((a+b+c+d+∣a+b−c−d∣)/2)  ≤ ((a+b+c+d+∣a−c∣+∣b−d∣)/2)  = ((a+c+∣a−c∣)/2)+((b+d+∣b−d∣)/2)  max(a+b,c+d) ≤ max(a,c)+max(b,d)
$$\mathrm{Inegalite}\:\mathrm{triangulaire}\:: \\ $$$$\mid{a}+{b}−{c}−{d}\mid\:\leqslant\:\mid{a}−{c}\mid+\mid{b}−{d}\mid \\ $$$$\Rightarrow\:\frac{\mid{a}+{b}−{c}−{d}\mid}{\mathrm{2}}\:\leqslant\:\frac{\mid{a}−{c}\mid+\mid{b}−{d}\mid}{\mathrm{2}} \\ $$$$\Rightarrow\:\frac{{a}+{b}+{c}+{d}+\mid{a}+{b}−{c}−{d}\mid}{\mathrm{2}} \\ $$$$\leqslant\:\frac{{a}+{b}+{c}+{d}+\mid{a}−{c}\mid+\mid{b}−{d}\mid}{\mathrm{2}} \\ $$$$=\:\frac{{a}+{c}+\mid{a}−{c}\mid}{\mathrm{2}}+\frac{{b}+{d}+\mid{b}−{d}\mid}{\mathrm{2}} \\ $$$$\mathrm{max}\left({a}+{b},{c}+{d}\right)\:\leqslant\:\mathrm{max}\left({a},{c}\right)+\mathrm{max}\left({b},{d}\right) \\ $$
Answered by Olaf_Thorendsen last updated on 15/Jul/21
Differentes approches mais une   approche elegante consiste a remarquer  que max(x,y) = ((x+y+∣x−y∣)/2).  Il ne reste alors qu′a utiliser  l′inegalite triangulaire.
$$\mathrm{Differentes}\:\mathrm{approches}\:\mathrm{mais}\:\mathrm{une}\: \\ $$$$\mathrm{approche}\:\mathrm{elegante}\:\mathrm{consiste}\:\mathrm{a}\:\mathrm{remarquer} \\ $$$$\mathrm{que}\:\mathrm{max}\left({x},{y}\right)\:=\:\frac{{x}+{y}+\mid{x}−{y}\mid}{\mathrm{2}}. \\ $$$$\mathrm{Il}\:\mathrm{ne}\:\mathrm{reste}\:\mathrm{alors}\:\mathrm{qu}'\mathrm{a}\:\mathrm{utiliser} \\ $$$$\mathrm{l}'\mathrm{inegalite}\:\mathrm{triangulaire}. \\ $$

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