Suppose f is a concave function on the interval [ a , b ], meaning λ f ( x ) + ( 1 − λ ) f ( y ) ≤ f ( λ x + ( 1 − λ ) y )for every x , y ∈ [ a , b ] and every λ ∈ [ 0 , 1 ]. Prove that for any p , q , r ∈ [ a , b ] with q ≥ r ≥ 0

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Suppose   f is a concave function on the interval   [  a  ,  b  ], meaning  λ  f  (  x  )  +  (  1  −  λ  )  f  (  y  )  ≤  f  (  λ  x  +  (  1  −  λ  )  y  )for every   x  ,  y  ∈  [  a  ,  b  ] and every   λ  ∈  [  0  ,  1  ]. Prove that for any   p  ,  q  ,  r  ∈  [  a  ,  b  ] with   q  ≥  r  ≥  0

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Let&#039;s calculate first the specific   λ that &quot;averages&quot; the endpoints   p  ±  q to give   p  ±  r, i.e. solve for   λ,                    λ        (        p        +        q        )        +        (        1        −        λ        )        (        p        −        q        )        =        ±        r                            (1)                            ⟹                          λ                      ±                          =                              ±            r            +            q                                2            q                              Now, using the fact that   q  ≥  r  ≥  0    ⟹    p  ±  r  ∈  [  p  −  q  ,  p  +  q  ],                    f        (        p        ±        r        )                            =        f        (                  λ          ±                (        p        +        q        )        +        (        1        −                  λ          ±                )        (        p        −        q        )        )                                          ≥                  λ          ±                f        (        p        +        q        )        +        (        1        −                  λ          ±                )        f        (        p        −        q        )            Summing these and substituting the values in   (  1  ),                    f        (        p        +        r        )        +        f        (        p        −        r        )                            ≥        (                  λ          +                +                  λ          −                )        f        (        p        +        q        )        +        (        2        −                  λ          +                −                  λ          −                )        f        (        p        −        q        )                                          =        f        (        p        +        q        )        +        f        (        p        −        q        )

Suppose f is a concave function on the interval [a,b], meaning lambda f(x)+(1−lambda)f(y)<=f(lambda x+(1−lambda)y) for every x,y in [a,b] and every lambda in [0,1]. Prove that for any p,q,r in a,b] with q>=r>=0

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New Questions in Calculus 1