Prove the following statement, in generic form, for f &gt; 0 - concave function: A f ( x A ) &gt; a 1 f ( x a 1 ) + a 2 f ( x a 2 ) + ⋯ + a N f ( x a N ) ,where x &gt; 0 and: ∑ i = 1 N a i = A .

Question

Prove the following statement, in generic form, for   f  &amp;gt;  0 - concave function:  A  f      (          x      A        )    &amp;gt;      a    1    f      (          x              a        1              )    +      a    2    f      (          x              a        2              )    +  ⋯  +      a    N    f      (          x              a        N              )    ,where   x  &amp;gt;  0 and:      ∑          i      =      1        N        a    i    =  A  .

luthersavage6lm · Accepted Answer

For example, we are trying to prove that   f  (  x  )  &amp;gt;      1    2    f  (  2  x  )  +      1    3    f  (  3  x  )  +      1    6    f  (  6  x  ), with       1    2    +      1    3    +      1    6    =  1.Take   f  (  x  )  =      1          1      −      x        +  C (with a big enough   C), which is concave on (  (  1  ,  +  ∞  )), and let   x  →  1  +  0. The LHS diverges (to   −  ∞) while the RHS converges (to  −      3    10  ) so the inequality must be broken for x close enough to   1.Start with   C  =  0 and then, once such x is found so that the inequality is broken, increase   C to get to another counterexample, where the function is also positive on   [  x  ,  ∞  ).

Prove the following statement, in generic form, for f>0 - concave function: Af(x/A)>a_1f(x/a_1)+a_2f(x/a_2)+⋯+a_Nf(x/a_N), where x>0 and: sum_(i=1)^N ai=A.

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