If a and b are positive numbers, prove that the equation 1 x − a + 1 x − b + 1 x = 0has two real roots; one between 1 3 a and 2 3 a , and one between − 2 3 b and − 1 3 b .

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If a and b are positive numbers, prove that the equation      1          x      −      a        +      1          x      −      b        +      1    x    =  0has two real roots; one between       1    3    a and       2    3    a , and one between   −      2    3    b and   −      1    3    b .

extractumzz · Accepted Answer

Step 1  f  (  x  )  =      1          x      −      a        +      1          x      +      b        +      1    x    =  0           a  ,  b  &amp;gt;  0.           (  1  )Then  f  (  a      /    3  )  =            9      (      a      +      b      )              2      a      (      a      +      3      b      )        &amp;gt;  0and  f  (  2  a      /    3  )  =     −            9      b              2      a      (      2      a      +      3      b      )        &amp;lt;  0. .So by the Intermediate Value Theorem (IVT) this equation (1) has one root in   (  a      /    2  ,  2  a      /    3  )  . . Next,  f  (  −  b      /    3  )  =     −            9      (      a      +      b      )              2      b      (      3      a      +      b      )        &amp;lt;  0and  f  (  −  2  b      /    3  )  =            9      a              2      b      (      3      a      +      2      b      )        &amp;gt;  0. .So again by IVT, equation (1) has one root in   (  −  2  b      /    3  ,  −  b      /    3  )  . .Hence proved.

Laylah Mora · Accepted Answer

Step 1  4  (      a          2        −  a  b  +      b          2        )  =  (  2  a  −  b      )          2        +  3      b          2      I think question is wrong. For exmaple, if   a  =  1  ,  b  =  2 , then two solutions are   1  ±      1          3       , and none of them lies between   −  4      /    3 and   −  2      /    3 since both are positive.

If a and b are positive numbers, prove that the equation 1 x −

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