another inequality 1 x y + z + 1 y z + x + 1 z x + y ≤ 1 2 Let x , y , z &gt; 0 and such x y z ≥ 2 + x + y + z, show that (1) 1 x y + z + 1 y z + x + 1 z x + y ≤ 1 2 The theory basis of speculation Use the following classical results x y z = 2 + x + y + z ⟹ x y + y z + z x ≥ 2 ( x + y + z )

Question

another inequality       1          x      y      +      z        +      1          y      z      +      x        +      1          z      x      +      y        ≤      1    2  Let   x  ,  y  ,  z  &amp;gt;  0 and such   x  y  z  ≥  2  +  x  +  y  +  z, show that                    (1)                                          1                          x              y              +              z                                      +                              1                          y              z              +              x                                      +                              1                          z              x              +              y                                      ≤                              1            2                              The theory basis of speculation Use the following classical results  x  y  z  =  2  +  x  +  y  +  z  ⟹  x  y  +  y  z  +  z  x  ≥  2  (  x  +  y  +  z  )

Kaya Kemp · Accepted Answer

I agree with you. Your inequality is true!Indeed, the condition gives       ∑          c      y      c            1          x      +      1        ≤  1Let   x  =  a,   y  =  b and   z  =  k  c, where   k  &amp;gt;  0 and       ∑          c      y      c            1          a      +      1        =  1. Hence,      1          a      +      1        +      1          b      +      1        +      1          c      +      1        ≥      1          a      +      1        +      1          b      +      1        +      1          k      c      +      1      which gives   k  ≥  1Thus,       ∑          c      y      c            1          x      y      +      z        =      1          a      b      +      k      c        +      1          k      a      c      +      b        +      1          k      b      c      +      a        ≤      1          a      b      +      c        +      1          a      c      +      b        +      1          b      c      +      a      Id est, it remains to prove that       ∑          c      y      c            1          a      b      +      c        ≤      1    2   for positives   a,   b and   c such that       ∑          c      y      c            1          a      +      1        =  1Let   a  =            y      +      z        x   and   b  =            x      +      z        y  , where   x,   y and   z are positive numbers.Hence,   c  =            x      +      y        z   and we need to prove that       ∑          c      y      c            1                            x          +          y                z            ⋅                        x          +          z                y            +                        y          +          z                x              ≤      1    2   or      ∑          c      y      c                  x      y      z              x      (      x      +      y      )      (      x      +      z      )      +      y      z      (      y      +      z      )        ≤      1    2  or      ∑          c      y      c                  x      y      z              (      x      +      y      +      z      )      (              x        2            +      y      z      )        ≤      1    2  or      ∑          c      y      c            (          x      2        −                  x        y        z                              x          2                +        y        z              )    ≥  0or      ∑          c      y      c                  x      (              x        2            −      y      z      )                      x        2            +      y      z        ≥  0or      ∑          c      y      c                  x      (      (      x      −      y      )      (      x      +      z      )      −      (      z      −      x      )      (      x      +      y      )      )                      x        2            +      y      z        ≥  0or      ∑          c      y      c        (  x  −  y  )      (                  x        (        x        +        z        )                              x          2                +        y        z              −                  y        (        y        +        z        )                              y          2                +        x        z              )    ≥  0or      ∑          c      y      c        z  (  x  −  y      )    2    (      x    2    +      y    2    +  x  z  +  y  z  )  (      z    2    +  x  y  )  ≥  0Done!

another inequality 1 x y + z </mrow> </mfrac>

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