Question 1209908
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Let w, x, y, and z be positive real numbers. If w + 2x + 3y + 6z = 8 - w^2 - x^2 - y^2 - z^2, 
then what is the maximum value of wxyz?
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The solution in the post by @CPhill is INCORRECT.


The right way to solve was presented by the artificial intelligence (Google AI) under this link


https://www.google.com/search?q=Let+w%2C+x%2C+y%2C+and+z+be+positive+real+numbers.+If+w+%2B+2x+%2B+3y+%2B+6z+%3D+8+-+w%5E2+-+x%5E2+-+y%5E2+-+z%5E2%2C+then+what+is+the+maximum+value+of+wxyz%3F&rlz=1C1CHBF_enUS1071US1071&oq=Let+w%2C+x%2C+y%2C+and+z+be+positive+real+numbers.+If+w+%2B+2x+%2B+3y+%2B+6z+%3D+8+-+w%5E2+-+x%5E2+-+y%5E2+-+z%5E2%2C+then+what+is+the+maximum+value+of+wxyz%3F&gs_lcrp=EgZjaHJvbWUyBggAEEUYOdIBCTE2OTFqMGoxNagCCLACAfEF_AIV2ztMxXzxBfwCFds7TMV8&sourceid=chrome&ie=UTF-8


but this solution has arithmetic error on the way, making the AI answer incorrect 
(so, the AI implementation was not precisely accurate).


I fixed this error, and now I am placing my CORRECT solution below.



<pre>
On the way, I will use the AM-GM inequality, which says that for non-negative real numbers {{{a[1]}}}, {{{a[2]}}}, . . . , {{{a[n]}}}


    {{{(a[1] + a[2] + ellipsis + a[n])/n}}} >= {{{root(n,a[1]*a[2] * ellipsis * a[n])}}}



Step 1

Rearrange the equation

    w^2 + w + x^2 + 2x + y^2 + 3y + z^2 + 6z = 8.


Step 2

Complete the squares

    Complete the squares for each variable

        (w^2 + w + 1/4) + (x^2 + 2x + 1) + (y^2 + 3y + 9/4) + (z^2 + 6z + 9) = 8 + 1/4 + 1 + 9/4 + 9

        (w+1/2)^2 + (x+1)^2 + (y+3/2)^2 + (z+3)^2 = 41/2


Step 3

Apply AM-GM inequality to the terms  (w+1/2)^2,  (x+1)^2,  (y+3/2)^2  and  (z+3)^2 :

    {{{((w+1/2)^2+(x+1)^2 + (y+3/2)^2+(z+3)^2)/4)}}} >= {{{root(4,(w+1/2)^2*(x+1)^2*(y+3/2)^2*(z+3)^2)}}},

    {{{41/8}}} >= {{{sqrt((w+1/2)*(x+1)*(y+3/2)*(z+3))}}}

    {{{(41/8)^2}}} >= {{{(w+1/2)*(x+1)*(y+3/2)*(z+3)}}}


Step 4

Apply AM=GM inequality to w and 1/2;  to x and 1;  to y and 3/2;  to z and 3:

    {{{w+1/2}}} >= {{{2*sqrt(w/2)}}}

    x + 1 >= {{{2*sqrt(x)}}}

    {{{y + 3/2}}} >= {{{2*sqrt((3y)/2)}}}

    {{{z+3}}} >= {{{2*sqrt(3z)}}}


Step 5

Combine the inequalities

    Multiply the inequalities

        {{{(w+1/2)*(x+1)*(y+3/2)*(z+3)}}} >= {{{(2*sqrt(w/2))*(2*sqrt(x))*(2*sqrt((3y)/2))*(2*sqrt(3x))}}}

        {{{(w+1/2)*(x+1)*(y+3/2)*(z+3)}}} >= {{{16*sqrt((9wxyz)/4))}}}

        {{{(w+1/2)*(x+1)*(y+3/2)*(z+3)}}} >= {{{24*sqrt(wxyz)}}}



Step 6

Find the maximum value of wxyz

    Combine the inequalities from  Step 3  and  Step 5

        {{{(41/8)^2}}} >= {{{24*sqrt(wxyz)}}}

        {{{1681/64}}} >= {{{24*sqrt(wxyz)}}}

        {{{1681/(64*24)}}} >= {{{sqrt(wxyz)}}}

        {{{1681/1536}}} >= {{{sqrt(wxyz)}}}

        {{{(1681/1536)^2}}} >= wxyz

        wxyz <= {{{(1681/1536)^2}}} = 1.1977136  (rounded).


<U>ANSWER</U>.  wxyz <= {{{(1681/1536)^2}}} = 1.19772  (rounded up).
</pre>

Solved.