Not a harmonic mean question

Since $x+y+z=1$

So we need to find the maximum possible value of

$x(1-z)^{2}(1-x)^{3}(1-y)^{4}$

Apply $AM \geq GM$ in the following way.

$\dfrac{x+2(1-z)+3(1-x)+4\dfrac{(1-y)}{2}}{10}\geq\sqrt [ 10 ]{\dfrac{x(1-z)^{2}(1-x)^{3}(1-y)^{4}}{16} }$

$\dfrac{7-2(x+y+z)}{10} \geq \sqrt [ 10 ]{\dfrac{x(1-z)^{2}(1-x)^{3}(1-y)^{4}}{16} }$

This finally gives $x(1-z)^{2}(1-x)^{3}(1-y)^{4} \leq 2^{4}\times(\dfrac{1}{2})^{10}$

So the max value of $x(x+y)^{2}(y+z)^{3}(x+z)^{4}$ is $\dfrac{1}{2^{6}}=\dfrac{1}{64}$

Equality holds when $x=z=\dfrac{1}{2}$ and $y=0$

I had a different approach...

First note that

$x+y=1-z$

$y+z=1-x$

$x+z=1-y$ .

Then let

$f(x,y,z)=x(1-z)^2(1-x)^3(1-y)^4$

If $x$ , $y$ , and $z$ are optimized such that $f(x,y,z)$ is maximized, then we claim $y$ is the least of the variables.

If $z\leq y$ , then $1-y\leq1-z$ . Thus,

$f(x,y,z) \leq f(x,z,y)$ .

So interchanging $y$ and $z$ yields a greater solutions, so we may assume $y \leq z$ .

Similarly, if $x\leq y$ , then $1-y\leq 1-x$ . Thus,

$f(x,y,z) \leq f(y,x,z)$ .

So interchanging $x$ and $y$ yields a greater solution, so again, we may assume $y\leq x$ .

Now to optimize $f$ , clearly we can choose $y=0$ . Thus, since $x+y+z=x+z=1$ , we can replace $1-z$ in $f$ with $x$ , giving us

$f(x)=x^3(1-x)^3$ .

We then use calculus. We have that

$f'(x)=3x^2(1-x)^3-3x^3(1-x)^2=3x^2(1-x)^2(1-2x)$

which has critical points $0$ , $1/2$ , and $1$ . Since $f(1)=f(0)=0$ , the maximized value is $f(1/2)=1/64$ .

Let $z=1-x-y$ , take the partial derivatives of the function, set them equal to 0, solve one of them for $y$ , plug in why into they other equation to get $x$ , use $x$ to find $y$ and $z$ , and finally plug them into the original equation to get $\frac{1}{64}$ , giving us our final answer of $p + q = 1 + 64 = 65$

P.S. (Sorry if the picture is sideways to you)

Fantastic solutions above. I went about this in a little bit different way. From the equation we know the biggest x+y, y+z, or x+z can be is 1. If one of these terms is 1 then the others must be less than one. Since these terms are positive and less than one the higher their order the lower their value. To maximize our value we should make x+z=1, which implies y=0. Simplifying the expression leaves us with (xz)^3 to maximize or just xz. Solving in terms of x gives us x-x^2. Finding the maxima of this give us x = 1/2 which means z = 1/2. Substituting gives us (1/2)^6 = 1/64. Thus p+q = 65.

Use A. M. - G. M. on x, x + y, x + y, y + z , y + z, y + z, and 0.5 x + 0.5 y four times. A. M. of these 10 is 0.5 which is the maximum value of tenth root of the product. Simply solving this gives the required expression to have a maximum value 1/64.

I decided that since the 4th power is the highest so I should maximize that one by making x+z=1 and y=0, then the expression becomes (x^3)(z^3), given x+z=1, I know from past experience, is maximized by setting x=1/2. You can prove this by inserting z=1-x and differentiating. So we have (1/2)^6

shubhendra singh solution - actually there are many ways to choose a,b,c,d such that a x+b (1-z)+c (1-x)+d(1-z)=k (x+y+z). But the important factor lies in the satisfiability of the equation which can be acheived only by checking equality condition. Here goes more rigorous proof of how to get to the values of a=1,b=2,c=3,d=2. Firstly the coefficients of x,y,z in the LHS of the above stated equation must be equal. This gives b=d and a+b=c. Now the other condition is that the AM-GM inequality must be satisfiable. So, we get 3 equations which must be consistent . They are :- a x=(b/2) (x+y) - (eq1) a x=(d/4) (x+z) - (eq2) a x=(c/3) (y+z) - (eq3) Finding values of y and z in terms of x from eq1 and eq2 and substituting in eq3 leads us to (2 a)/b+(4 a)/d=2+(3 a)/c Now putting b=d we get (6 a)/b=2+(3 a)/c Now putting a=c-b we get 6 c c-11 b c+3 b b=0 Solving this we get b=3 c and 2 c=3 b b=3 c could be ruled out as substituting this in a+b=c leads to a=-2 c which means either a or c is negative which is not desirable. So we take 2 c=3 b so we put c=3, b=2 From these two values d=2 as (b=d) and a=1 as (a=c-b) So this is how we could arrive at correct values of a,b,c,d which fit into all the conditions

Why are the solutions dividing by 2 when turning x+z into 1-y? The AM should be [x + 2(1-z) + 3(1-x) + 4(1-y)] / 10 based on the last term being (x+z)^4 right?