Prove that 1 + a b 1 + a + 1 + b c 1 + b + 1 + c a 1 + c ≥ 3I have to prove that 1 + a b 1 + a + 1 + b c 1 + b + 1 + c a 1 + c ≥ 3is always true for real numbers a , b , c &gt; 0 with a b c = 1Using the AM-GM inequality I got as far as 1 + a b 1 + a + 1 + b c 1 + b + 1 + c a 1 + c ≥ b b a + 1 + c c b + 1 + a a c + 1 but I do not yet know how to finish my proof from there (if this is helpful at all?!).

Question

Prove that             1      +      a      b              1      +      a        +            1      +      b      c              1      +      b        +            1      +      c      a              1      +      c        ≥  3I have to prove that            1      +      a      b              1      +      a        +            1      +      b      c              1      +      b        +            1      +      c      a              1      +      c        ≥  3is always true for real numbers   a  ,  b  ,  c  &amp;gt;  0 with   a  b  c  =  1Using the AM-GM inequality I got as far as            1      +      a      b              1      +      a        +            1      +      b      c              1      +      b        +            1      +      c      a              1      +      c        ≥      b          b              a        +        1              +      c          c              b        +        1              +      a          a              c        +        1            but I do not yet know how to finish my proof from there (if this is helpful at all?!).

vibrerentb · Accepted Answer

Hint:             1      +      a      b              1      +      a        =            a      b      c      +      a      b              1      +      a        =  a  b            (                  1      +      c              1      +      a                  )      ?

musicbachv7 · Accepted Answer

Now I see. Using Especially Lime&#039;s comment, we can see that      (                  1        +        a        b                    1        +        a              )        (                  1        +        b        c                    1        +        b              )        (                  1        +        c        a                    1        +        c              )    =  1.Hence, we can set  x  :=            1      +      a      b              1      +      a        ,  y  :=            1      +      b      c              1      +      b      and our inequality becomes  x  +  y  +      1          x      y        ≥  3which is equivalent to            x      +      y      +              1                  x          y                      3    ≥  1.But using the AM-GM inequality for three variables, we can see that            x      +      y      +              1                  x          y                      3    ≥            x      y      ⋅              1                  x          y                      3    =  1  ,and our claim follows.

I have to prove that (1+ab)/(1+a)+(1+bc)/(1+b)+(1+ca)/(1+c)>=3 is always true for real numbers a,b,c>0 with abc=1.

Open question

Answer & Explanation

New Questions in Pre-Algebra