How to show simple inequality of fractionsIf a a + b &lt; a ′ a ′ + b ′ then how can I show that a a + 2 b &lt; a ′ a ′ + 2 b ′ ∀ a , b , c &gt; 0I tried puitting in a constant k so a a + b = k ∗ a a + 2 b and I get k = a + b a + 2 b and I dont know where from there.

Question

How to show simple inequality of fractionsIf      a          a      +      b        &amp;lt;            a      ′                      a        ′            +              b        ′            then how can I show that      a          a      +      2      b        &amp;lt;            a      ′                      a        ′            +      2              b        ′                 ∀     a  ,  b  ,  c  &amp;gt;  0I tried puitting in a constant k so      a          a      +      b        =  k  ∗      a          a      +      2      b      and I get  k  =            a      +      b              a      +      2      b      and I dont know where from there.

Brennen Bishop · Accepted Answer

Looks like it isn&#039;t true even for all numbers positive.      30    6    &amp;lt;      6    1  but      130    106    &amp;gt;      106    101

Trevor Wood · Accepted Answer

I&#039;ll assume everything is positive.If       a          a      +      b        &amp;lt;            a      ′                      a        ′            +              b        ′             then   a  (      a    ′    +      b    ′    )  &amp;lt;      a    ′    (  a  +  b  ) We want to know when       a          a      +      2      b        &amp;lt;            a      ′                      a        ′            +      2              b        ′             or   a  (      a    ′    +  2      b    ′    )  &amp;lt;      a    ′    (  a  +  2  b  )Let   u  =      a    ′    (  a  +  b  )  −  a  (      a    ′    +      b    ′    )  &amp;gt;  0 and   v  =      a    ′    (  a  +  2  b  )  −  a  (      a    ′    +  2      b    ′    ). We want to when   v  &amp;gt;  0Since   u  =      a    ′    b  −  a      b    ′    &amp;gt;  0,       a    ′    b  &amp;gt;  a      b    ′  We have   v  =  (      a    ′    (  a  +  b  )  +      a    ′    b  )  −  (  a  (      a    ′    +      b    ′    )  +  a      b    ′    )  =      a    ′    (  a  +  b  )  −  a  (      a    ′    +      b    ′    )  +      a    ′    b  −  a      b    ′    =  u  +      a    ′    b  −  a      b    ′  Since, as shown above,       a    ′    b  &amp;gt;  a      b    ′   and    u  &amp;gt;  0, then   v  &amp;gt;  0 as we wanted to show.

How to show simple inequality of fractions If a a + b </

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