Proving that 1 3 x 2 + 1 + 1 3 y 2 + 1 + 1 3 z 2 + 1 ≥ 3 16 Let x , y , z ≥ 1 and x + y + z = 6. Then 1 3 x 2 + 1 + 1 3 y 2 + 1 + 1 3 z 2 + 1 ≥ 3 16 .I tried to use Cauchy- Schwartz inequality but it doesn't work.

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Proving that       1          3              x        2            +      1        +      1          3              y        2            +      1        +      1          3              z        2            +      1        ≥      3    16  Let   x  ,  y  ,  z  ≥  1 and   x  +  y  +  z  =  6. Then      1          3              x        2            +      1        +      1          3              y        2            +      1        +      1          3              z        2            +      1        ≥      3    16    .I tried to use Cauchy- Schwartz inequality but it doesn&#039;t work.

faliryr · Accepted Answer

Also, the Tangent Line method works:      ∑          c      y      c            1          3              x        2            +      1        =      ∑          c      y      c            (          1              3                  x          2                +        1              −          1      13        +          12      169        (    x    −    2    )    )    +      3    13    =  =      ∑          c      y      c                  3      (      x      −      2              )        2            (      12      x      +      11      )              169      (      3              x        2            +      1      )        +      3    13    ≥      3    13    &amp;gt;      3    16    .

excefebraxp · Accepted Answer

Using Lagrange multipliers, write  F  (  x  ,  y  ,  z  )  =      1          3              x        2            +      1        +      1          3              y        2            +      1        +      1          3              y        2            +      1        −  λ  (  x  +  y  +  z  )Taking derivatives, we get            6      x              (      3              x        2            +      1              )        2              =            6      y              (      3              y        2            +      1              )        2              =            6      z              (      3              z        2            +      1              )        2              =  −  λNow it&#039;s straightforward to show that   f  (  t  )  =            6      t              (      3              t        2            +      1              )        2             is one-to-one on   t  &amp;gt;  1  , so   x  =  y  =  z  =  2.

Proving that 1/(3x^2+1)+1/(3y^2+1)+1/(3z^2+1)>= 3/16 Let x,y,z>=1 and x+y+z=6. I tried to use Cauchy- Schwartz inequality but it doesn't work.

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