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Question Number 193137 by Mastermind last updated on 04/Jun/23
1) Prove that:  ∣a+b+c∣≥∣a∣−∣b∣−∣c∣    2) Find all x∈R that satify the follow−  ing inequalities   i) ∣x^2 −4∣<5  ii) ∣x∣+∣x+2∣<5    Help!
1)Provethat:a+b+c∣⩾∣abc2)FindallxRthatsatifythefollowinginequalitiesi)x24∣<5ii)x+x+2∣<5Help!
Answered by aba last updated on 04/Jun/23
∣a∣=∣a+b+c−(b+c)∣≤∣a+b+c∣+∣b+c∣ (1)  ∣b+c∣≤∣b∣+∣c∣ ⇒ −∣b∣−∣c∣≤−∣b+c∣ (2)  (1)+(2)⇒∣a∣−∣b∣−∣c∣≤∣a+b+c∣ ✓
a∣=∣a+b+c(b+c)∣⩽∣a+b+c+b+c(1)b+c∣⩽∣b+cbc∣⩽b+c(2)(1)+(2)⇒∣abc∣⩽∣a+b+c
Answered by Rajpurohith last updated on 06/Jun/23
1)0≤∣a∣=∣a+b+c−b−c∣≤∣a+b+c∣+∣b∣+∣c∣  ⇒∣a∣−∣b∣−∣c∣≤∣a+b+c∣  2)  (i)Two cases arise   •x^2 ≤4  ⇒∣x^2 −4∣=4−x^2 <5 ⇒−1≤x^2  which is always true.  so x^2 ≤4 is feasible i.e, ∣x∣≤2.  •x^2 >4 ⇒∣x^2 −4∣=x^2 −4<5  ⇒x^2 <9 ⇒∣x∣<3  hence the final region  or solution is ∣x∣<3.    (ii) •Suppose x≥0  then ∣x∣=x and ∣x+2∣=x+2  so the inequality becomes  x+x+2<5  ⇒2x+2<5⇒x<(3/(2 )) . Hence 0≤x<(3/2) is a region.  •suppose −2≤x<0 ⇒x+2≥0  ⇒∣x∣=−x and ∣x+2∣=x+2  so the inequality becomes −x+x+2<5 which is true.  so −2≤x<0  is also a feasible region.  •suppose x<−2 ⇒∣x+2∣=−x−2 and ∣x∣=−x  so the inequality becomes  −x−x−2<5 ⇒−2x<7  ⇒x>−(7/2)  hence the feasible region of the inequality is  −(7/2)<x<(3/2)  .
1)0⩽∣a∣=∣a+b+cbc∣⩽∣a+b+c+b+c⇒∣abc∣⩽∣a+b+c2)(i)Twocasesarisex24⇒∣x24∣=4x2<51x2whichisalwaystrue.sox24isfeasiblei.e,x∣⩽2.x2>4⇒∣x24∣=x24<5x2<9⇒∣x∣<3hencethefinalregionorsolutionisx∣<3.(ii)Supposex0thenx∣=xandx+2∣=x+2sotheinequalitybecomesx+x+2<52x+2<5x<32.Hence0x<32isaregion.suppose2x<0x+20⇒∣x∣=xandx+2∣=x+2sotheinequalitybecomesx+x+2<5whichistrue.so2x<0isalsoafeasibleregion.supposex<2⇒∣x+2∣=x2andx∣=xsotheinequalitybecomesxx2<52x<7x>72hencethefeasibleregionoftheinequalityis72<x<32.

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