Proving inequality using Lagrange multipliers.I started learing about Lagrange Multipliers and I got the following question:Prove that a b + c + b c + a + c a + b ≥ 3 2 , for each a , b , c &gt; 0I'm not sure how to use Lagrange multipliers for it... any ideas?

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Proving inequality using Lagrange multipliers.I started learing about Lagrange Multipliers and I got the following question:Prove that       a          b      +      c        +      b          c      +      a        +      c          a      +      b        ≥      3    2  , for each   a  ,  b  ,  c  &amp;gt;  0I&#039;m not sure how to use Lagrange multipliers for it... any ideas?

Branson Grimes · Accepted Answer

Since our inequality is homogeneous, we can assume   a  +  b  +  c  =  3Also let   f  (  a  ,  b  ,  c  ,  λ  )  =      ∑          c      y      c            a          3      −      a        +  λ  (  a  +  b  +  c  −  3  )  −      3    2              ∂      f              ∂      a        =      3          (      3      −      a              )        2              +  λLet in   (  a  ,  b  ,  c  ) our   f gets a minimal value (  f is a continuous function and this thing happens on the compact).Thus, in this point             ∂      f              ∂      a        =            ∂      f              ∂      b        =            ∂      f              ∂      c        =  0, which gives      3          (      3      −      a              )        2              =      3          (      3      −      b              )        2            and since   a  +  b  &amp;lt;  3, we obtain:   a  =  b, which says that   c  =  3  −  2  a and it remains to prove that            2      a              3      −      a        +            3      −      2      a              2      a        ≥      3    2    ,where   0  &amp;lt;  a  &amp;lt;      3    2  , which gives   (  a  −  1      )    2    ≥  0Also for   a  →      0    +  ... our inequality is obviously true.Done!

Memphis Khan · Accepted Answer

Let   f  (  x  ,  y  ,  z  )  =      x          y      +      z        +      y          x      +      z        +      z          x      +      y                                                ∇            →                          f        ⋅                              x            ^                                              =                    +                  1                      (            y            +            z            )                          −                  y                      (            x            +            z                          )              2                                      −                  z                      (            x            +            y                          )              2                                                                                ∇            →                          f        ⋅                              y            ^                                              =                    −                  x                      (            y            +            z                          )              2                                      +                  y                      (            x            +            z            )                          −                  z                      (            x            +            y                          )              2                                                                                ∇            →                          f        ⋅                              z            ^                                              =                    −                  x                      (            y            +            z                          )              2                                      −                  y                      (            x            +            z                          )              2                                      +                  z                      (            x            +            y            )                              You should be able to demonstrate that            ∇      →        f  =  0    ⟹    x  =  y  =  zand since   f  (  x  ,  x  ,  x  )  =      3    2   you are pretty well done.

Proving inequality using Lagrange multipliers. I started learing about Lagrange Multipliers and I got the following question: Prove that a/(b+c)+b/(c+a)+c/(a+b)>=3/2, for each a,b,c>0.I'm not sure how to use Lagrange multipliers for it... any ideas?

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