Ratio InequalityHow can I prove that, a 1 + a 2 + ⋯ + a n b 1 + b 2 + ⋯ + b n ≤ max i { a i b i } where 1 ≤ i ≤ n, and a i ≠ a j and b i ≠ b j , ∀ i ≠ jEdit I have figured out that the above assumptions about a i , and b i are not needed.

Question

Ratio InequalityHow can I prove that,                    a                  1                    +              a                  2                    +      ⋯      +              a                  n                                    b                  1                    +              b                  2                    +      ⋯      +              b                  n                      ≤      max    i        {                  a                  i                            b                  i                      }  where   1  ≤  i  ≤  n, and       a          i        ≠      a          j       and       b          i        ≠      b          j        ,  ∀  i  ≠  jEdit I have figured out that the above assumptions about       a          i      , and       b          i       are not needed.

kuncwadi17 · Accepted Answer

The       a    i   can be arbitrary real numbers, but the       b    i   need to be positive. Then                    (*)                              a          j                ≤                  b          j                ⋅                  {                      max            i                                              a              i                                      b              i                                }                                  for                       j            =            1            ,            …            ,            n                              and adding these gives the desired inequality.If the       b    i   are not required to be positive then the inequality must not hold, a counter-example is            2      −      1              3      −      2        &amp;gt;  max      {          2      3        ,                  −        1                    −        2              }      .

Averi Mitchell · Accepted Answer

Suppose this holds for   n  =  2 (prove this base case yourself). Then            (              a        1            +      .      .      .              a        k            )      +              a                  k          +          1                            (              b        1            +      .      .      .      +              b        k            )      +              b                  k          +          1                      ≤  max      (                            a          1                +        .        .        .        +                  a          k                                      b          1                +        .        .        .        +                  b                      k                                ,                  a                  k          +          1                            b                  k          +          1                      )

Ratio Inequality How can I prove that, a <mrow class="MJX-Te

Answered question

Answer & Explanation

New Questions in Pre-Algebra