For real numbers 𝑎,𝑏, if 𝑎+𝑎𝑏+𝑏=3, then find the range of 𝑚=𝑎−𝑎𝑏+𝑏. Is there any inequalities here to use?

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Hamnetmj · Accepted Answer

Set   s  =  a  +  b and   t  =  a  −  b. This is a lossless change of variables, so transform the condition:                    3                            =        a        +        b        +        a        b                            12                            =        4        s        +        (        s        +        t        )        (        s        −        t        )                                      t          2                                    =                  s          2                +        4        s        −        12        =        (        s        +        6        )        (        s        −        2        )            Therefore the condition can be satisfied only when the right side isn&#039;t negative, i.e.   s  ∉  (  −  6  ,  2  ).You want to find the range of   a  +  b  −  a  b  =  a  +  b  −  (  3  −  a  −  b  )  =  2  s  −  3, but that&#039;s clearly       R    ∖  (  −  15  ,  1  ).

Bodonimhk · Accepted Answer

Not sure if you can get a more elementary answer, but you could see   1  +  a  ,  1  +  b are roots of       x    2    −  s  x  +  4  =  0, where       s    2    ≥  16. So, we have   m  =  (  a  +  a  b  +  b  )  −  2  (  1  +  a  )  (  1  +  b  )  +  2  (  1  +  a  +  1  +  b  )  −  2  =  2  s  −  7  ∉  (  −  15  ,  1  ).

For real numbers 𝑎,𝑏, if 𝑎+𝑎𝑏+𝑏=3, then find the range of 𝑚=𝑎−𝑎𝑏+𝑏. Is there any inequalities here to use?

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