Show that ( a + 2 b + 2 a + 1 ) ( b + 2 a + 2 b + 1 ) ⩾ 16 if a b ⩾ 1 a, b are positive real numbers.Should I use AM-GM inequality? If yes, where?

Question

Show that   (  a  +  2  b  +      2          a      +      1        )  (  b  +  2  a  +      2          b      +      1        )  ⩾  16 if   a  b  ⩾  1  a,   b are positive real numbers.Should I use AM-GM inequality? If yes, where?

Yasmin Lam · Accepted Answer

Let   a  +  b  =  2  u.Thus, by AM-GM   u  ≥  1 and by C-S and AM-GM we obtain:      (    a    +    2    b    +          2              a        +        1              )        (    b    +    2    a    +          2              b        +        1              )    =  =      (    a    +    b    +    b    +          2              a        +        1              )        (    a    +    b    +    a    +          2              b        +        1              )    ≥  ≥            (      a      +      b      +              a        b            +              2                  (          a          +          1          )          (          b          +          1          )                    )        2    ≥  ≥            (      2      u      +      1      +              4                  a          +          1          +          b          +          1                    )        2    =            (                        u          +          1                2            +              2                  u          +          1                    +              3        2            u      +              1        2            )        2    ≥  ≥            (      2      +              3        2            +              1        2            )        2    =  16.Done!

metal1fc · Accepted Answer

Expand to get:  a  b  +  2      a    2    +            2      a              b      +      1        +  2      b    2    +  4  a  b  +            4      b              b      +      1        +            2      b              a      +      1        +            4      a              a      +      1        +      4          (      a      +      1      )      (      b      +      1      )        ≥  9  +            2              a        2            +      2      a      +      4      a      b      +      4      b      +      2              b        2            +      2      b      +      4      a      b      +      4      a      +      4              (      a      +      1      )      (      b      +      1      )        ≥  9  +  7  =  16The last inequality follows as:  2      a    2    +  2  a  +  4  a  b  +  4  b  +  2      b    2    +  2  b  +  4  a  b  +  4  a  +  4  ≥  7  (  a  +  1  )  (  b  +  1  )  2      a    2    +  2      b    2    +  a  b  ≥  a  +  b  +  3Now it&#039;s enough to prove that   2      a    2    +  2      b    2    ≥  a  +  b  +  2. From the condition we have by AM-GM   a  +  b  ≥  2      a    b    ≥  2. Hence by using the quadratic mean inequality:  2      a    2    +  2      b    2    ≥  (  a  +  b      )    2    ≥  2  (  a  +  b  )  ≥  a  +  b  +  2Hence the proof.

Show that (a+2b+(2)/(a+1))(b+2a+(2)/(b+1))>=16 a, b are positive real numbers. Should I use AM-GM inequality? If yes, where?

Answered question

Answer & Explanation

New Questions in Pre-Algebra