Let f be a concave function. Then, by definition, for any α ∈ [ 0 , 1 ] f ( α x + ( 1 − α ) y ) ≥ α f ( x ) + ( 1 − α ) f ( y )Is there a way to prove that f ( x ) + f ( y ) ≤ f ( α x + ( 1 − α ) y ) + f ( α y + ( 1 − α ) x )by using the upper definition?

Question

Let   f be a concave function. Then, by definition, for any   α  ∈  [  0  ,  1  ]  f  (  α  x  +  (  1  −  α  )  y  )  ≥  α  f  (  x  )  +  (  1  −  α  )  f  (  y  )Is there a way to prove that  f  (  x  )  +  f  (  y  )  ≤  f  (  α  x  +  (  1  −  α  )  y  )  +  f  (  α  y  +  (  1  −  α  )  x  )by using the upper definition?

Samsonitew7b · Accepted Answer

We have      {                            f          (          α          x          +          (          1          −          α          )          y          )                          ≥          α          f          (          x          )          +          (          1          −          α          )          f          (          y          )                                      f          (          (          1          −          α          )          x          +          α          y          )                          ≥          (          1          −          α          )          f          (          x          )          +          α          f          (          y          )                        By adding those two inequalities you get the desired result  f  (  α  x  +  (  1  −  α  )  y  )  +  f  (  α  y  +  (  1  −  α  )  x  )  ≥  f  (  x  )  +  f  (  y  )

Let f be a concave function. Then, by definition, for any α in [0,1] f(alpha x+(1−alpha)y)>=alpha f(x)+(1−alpha)f(y) Is there a way to prove that f(x)+f(y)>=f(alpha x(1−alpha )y)+f(alpha y+(1−alpha)x) by using the upper definition?

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New Questions in Calculus 1