If both a , b &gt; 0, then a a b b ≥ a b b a , if both a and b are positive.

Question

If both   a  ,  b  &amp;gt;  0, then       a    a        b    b    ≥      a    b        b    a  , if both a and b are positive.

ebizsavvy1txn · Accepted Answer

We just have to show that       a          a      −      b        ≥      b          a      −      b      . This is equivalent to   (      a    b        )          a      −      b        ≥  1.If   a  ≥  b, then       a    b    ≥  1, Also   a  −  b  ≥  0. A number greater than   1 raised to a positive exponent is clearly greater than   1.If   a  ≤  b, then       a    b    ≤  1.   a  −  b  ≤  0. A positive number less than   1 raised to a negative exponent is greater than   1.Hence we are done as we considered both cases.

inurbandojoa · Accepted Answer

log  ⁡  (      a    a        b    b    )  =  a  log  ⁡  a  +  b  log  ⁡  b  log  ⁡  (      a    b        b    a    )  =  a  log  ⁡  b  +  b  log  ⁡  aThus, by the rearrangement inequality, because   log is strictly increasing,  log  ⁡  (      a    a        b    b    )  ≥  log  ⁡  (      a    b        b    a    )Similarly, because   log is strictly increasing,      a    a        b    b    ≥      a    b        b    a    . This can be generalized as follows. Let       σ    1    ,      σ    2    ,  .  .  .  ,      σ    n   be any permutation of   1  ,  2  ,  .  .  .  ,  n, then      a    1                  a        1                  a    2                  a        2                  a    3                  a        3              ⋯      a    n                  a        n              ≥      a    1                  a                              σ            1                                    a    2                  a                              σ            2                                ⋯      a          n      −      1                      a                              σ                          n              −              1                                                a    n                  a                              σ            n                                ≥      a    1                  a        n                  a    2                  a                  n          −          1                      ⋯      a          n      −      1                      a        2                  a    n                  a        1            by an identical argument.

If both a,b>0, then a^ab^b >= a^bb^a, if both a and b are positive.

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