Show that 1 a + 1 b ≠ 1 a + b Problem Assume that a , b ∈ R − { 0 } and that a + b ≠ 0. Prove that 1 a + 1 b ≠ 1 a + b My ProofLet's assume that 1 a + 1 b = 1 a + b , then it follows that ( a + b ) 2 − a b = 0Let x = a + b and y = a b. Now b = x − a and so y = a ( x − a ) = a x − a 2 . The previous equation can be written as x 2 − y = 0Substituting y = a x − a 2 in this equation gives x 2 − a x + a 2 = 0The discriminant of this quadratic equation (in x) is − 3 a 2 &lt; 0 and therefore x = a + b has no real solution. This means a + b ∈ C and therefore either of a or b or both are not real but this contradicts our assumption that a,b are real numbers. Therefore by contradiction 1 a + 1 b ≠ 1 a + b My QuestionIs my proof correct? are there any alternative proofs?

Question

Show that       1    a    +      1    b    ≠      1          a      +      b      Problem Assume that   a  ,  b  ∈      R    −  {  0  } and that    a  +  b  ≠  0. Prove that       1    a    +      1    b    ≠      1          a      +      b      My ProofLet&#039;s assume that       1    a    +      1    b    =      1          a      +      b      , then it follows that  (  a  +  b      )    2    −  a  b  =  0Let   x  =  a  +  b and   y  =  a  b. Now   b  =  x  −  a and so   y  =  a  (  x  −  a  )  =  a  x  −      a    2  . The previous equation can be written as      x    2    −  y  =  0Substituting   y  =  a  x  −      a    2   in this equation gives      x    2    −  a  x  +      a    2    =  0The discriminant of this quadratic equation (in x)  is   −  3      a    2    &amp;lt;  0 and therefore   x  =  a  +  b has no real solution. This means   a  +  b  ∈      C   and therefore either of a or b or both are not real but this contradicts our assumption that a,b are real numbers. Therefore by contradiction       1    a    +      1    b    ≠      1          a      +      b      My QuestionIs my proof correct? are there any alternative proofs?

Jovan Wong · Accepted Answer

You have       a    2    +  a  b  +      b    2    =  0. Treating this as a quadratic equation in which the unknown quantity is a gives us this solution:  a  =            −      b      ±                        b          2                −        4                  b          2                      2    =            −      b      ±      b              −        3              2    =  b      (                  −        1        ±        i                  3                    2        )    .So if b is any complex number except 0  (e.g. let   b  =  1 and a is as given above, then      1    a    +      1    b    =      1          a      +      b        .But otherwise this last identity does not hold.Alternatively, one could just seek counterexamples. For example if   a  =  1 and   b  =  1 then      1    a    +      1    b    =  1  +  1  =  2    and        1          a      +      b        =      1    2      and    2  ≠      1    2    .

Tristian Velazquez · Accepted Answer

Assume w.n.l.g. that   a  &amp;gt;  0 and   b  ∈  (  −  a  ,  a  ]. Now, if   b  &amp;gt;  0      1    a    +      1    b    &amp;gt;      1    a    &amp;gt;      1          a      +      b        . On the other hand, if   b  &amp;lt;  0      1    a    +      1    b    ≤  0  &amp;lt;      1          a      +      b        .

Show that 1 a </mfrac> + 1 b </mfrac> ≠ 1 a

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