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Question Number 157436 by MathSh last updated on 23/Oct/21

if x;y;z≥0 then: 2^(x+y+z) + 2 ≥ 2^x + 2^y + 2^z

ifx;y;z⩾0then:2x+y+z+2⩾2x+2y+2z

Answered by Jamshidbek last updated on 23/Oct/21

Solution. 2^x =a≥1 2^y =b≥1 2^z =c≥1 abc+2≥a+b+c ⇒ abc−a+2−(b+c)≥0 abc−bc−a+1+bc−b+1−c≥0 bc(a−1)−(a−1)+b(c−1)−(c−1)≥0 (a−1)(bc−1)+(b−1)(c−1)≥0

Solution.2x=a⩾12y=b⩾12z=c⩾1abc+2⩾a+b+c⇒abc−a+2−(b+c)⩾0abc−bc−a+1+bc−b+1−c⩾0bc(a−1)−(a−1)+b(c−1)−(c−1)⩾0(a−1)(bc−1)+(b−1)(c−1)⩾0

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