Inequality Problem 1 x + 1 y + 1 z ≥ 9 x + y + z Could I please have help solving this inequality. I have tried but haven't made meaningful progress.If x,y and z are real numbers prove that: 1 x + 1 y + 1 z ≥ 9 x + y + z

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Inequality Problem       1    x    +      1    y    +      1    z    ≥      9          x      +      y      +      z      Could I please have help solving this inequality. I have tried but haven&#039;t made meaningful progress.If x,y and z are real numbers prove that:      1    x    +      1    y    +      1    z    ≥      9          x      +      y      +      z

fairymischiefv9 · Accepted Answer

Multiply your inequality by   x  y  z  (  x  +  y  +  z  ), you get :      y    2    z  +  y      z    2    +      x    2    z  +  x      z    2    +      x    2    y  +  x      y    2    ≥  6  x  y  zNow redivide by   x  y  z (maybe not a good idea to multiply initially :-) :      y    x    +      z    x    +      x    y    +      z    y    +      x    z    +      y    z    ≥  6As it is well know that for all   u  &amp;gt;  0,   u  +      1    u    ≥  2, you get your inequality for   x  ,  y  ,  z  &amp;gt;  0Now for the case   x  ,  y  ,  z non null reals, I don&#039;t know...

Livia Cardenas · Accepted Answer

Obviously we must assume that the numbers are positive, as otherwise the inequality doesn&#039;t always hold. Nevertheless you can prove this inequality by using Cauchy-Schwartz Inequality. We have:      (          1      x        +          1      y        +          1      z        )    (  x  +  y  +  z  )  ≥            (                        x          x                    +                        y          y                    +                        z          z                    )        2    =  9Now dividing by   (  x  +  y  +  z  ) we get the wanted inequality.

Could I please have help solving this inequality. I have tried but haven't made meaningful progress. If x,y and z are real numbers prove that: 1/x + 1/y +1/z >= (9)/(x+y+z)

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