If x + y + z = 12 and x 2 + y 2 + z 2 = 54 then prove that one has to be smaller or equal to 3 and one has to be bigger or equal than 5.

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If   x  +  y  +  z  =  12 and       x    2    +      y    2    +      z    2    =  54 then prove that one has to be smaller or equal to 3 and one has to be bigger or equal than 5.

cokeman206 · Accepted Answer

Assume first that x,y,z are all distinct. W.l.o.g. we can assume that x&amp;gt;y&amp;gt;z. Then x,y,z are the three zeros of f(a). By Rolle&#039;s theorem f′(a) must have a zero between z and y and another zero between y and x. But you know the zeros of f′(a), so this shows thatz&amp;lt;3

Ralzereep9h · Accepted Answer

We might as well require   x  ≥  y  ≥  z. If   z  &amp;gt;  3, let   w  =  3  −  z and we are thinking of w as small. The largest       x    2    +      y    2    +      z    2   can be is   (  6  −  2  w      )    2    +  2  (  3  +  w      )    2    =  54  −  12  w  +  6      w    2  w+6w2, so   z  &amp;lt;  3 If   x  &amp;lt;  5, Let   x  =  5  −  w, then the largest       x    2    +      y    2    +      z    2   can be is   2  (  5  −  w      )    2    +  (  2  +  2  w      )    2    =  54  −  16  w  +  6      w    2  . In fact there is a solution   5  ,  5  ,  2

If x+y+z=12 and x^2+y^2+z^2=54 then prove that one has to be smaller or equal to 3 and one has to be bigger or equal than 5.

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