Very Very Weird Inequality

If we consider $f(x,y,z) = xy^2 + yz^2 + 2xyz$ for $x,y,z \ge 0$ and $x+y+z=5$ , then

• $f(x,y,0) = xy^2$ is maximised over $x,y \ge 0$ , $x+y=5$ by maximising $(5-y)y^2$ over $0 \le y \le 5$ . The maximum occurs when $y=\tfrac{10}{3}\,,\,x=\tfrac53$ , and the corresponding value of $f$ is $\tfrac{500}{27}$ .
• $f(x,0,z) = 0$ .
• $f(0,y,z) = yz^2$ is maximised over $y,z \ge 0$ , $y+z=5$ when $y=\tfrac52\,,\,z=\tfrac{10}{3}$ , and the maximum value of $f$ is $\tfrac{500}{27}$ .

If we want to find other extrema, we can seek to find stationary points for the function with open domain $g(x,y) \; =\; f(x,y,5-x-y) \; = \; y^3 + (x-10)y^2 + (25-x^2)y \hspace{2cm} x,y > 0 \,,\, x+y < 5$ Since $\nabla g \; = \; \left(\begin{array}{c} y(y-2x) \\ 3y^2 + 2(x-10)y + 25-x^2 \end{array} \right)$ and hence $\nabla g = \mathbf{0}$ in this open domain when $y=2x$ and $\begin{aligned} 3y^2 + 2(x-10)y + 25 - x^2 & = 0 \\ 12x^2 + 4x(x-10) + 25-x^2 & = 0 \\ 5(3x-5)(x-1) & = 0 \end{aligned}$ If $x=\tfrac53$ then $y=\tfrac{10}{3}$ and then $x+y=5$ , so we are not in the open domain. Thus the only stationary point of $g$ in the open domain occurs when $x=1,y=2$ . The Hessian matrix of $g$ has positive determinant $20$ and negative trace $-10$ at $(1,2)$ , and hence the stationary point at $(1,2)$ is indeed a maximum.

Thus the maximum of $f(x,y,z)$ for $x,y,z \ge 0$ and $x+y+z=5$ does not occur on the boundary of this domain (the maximum of $f$ on the boundary is $\tfrac{500}{27}$ ), but is $\boxed{20}$ , occurring at $(1,2,2)$ .

Using Lagrange multiplier method is just match this question.

First set two function

$F=x+y+z=5$

$G=xy^2 + yz^2 + 2xyz$

Thus , compute the method.

$k \bigtriangledown F = \bigtriangledown G$

$\Rightarrow k < 1 , 1 , 1 > = < y^2 + 2yz , 2xy + z^2 + 2xz , 2yz + 2xy >$

So we can get three equation plus the initial equation :

$y^2 + 2yz = k \ldots (1)$

$2xy + z^2 + 2xz = k \ldots (2)$

$2yz + 2xy = k \ldots (3)$

$x + y + z = 5 \ldots (4)$

Start to solve these equations

By comparing (3) and (4) , we get

$10y-2y^2 = k \ldots (5)$

By comparing (1) and (5) , we get

$3y + 2z - 10 = 0 \ldots (6)$

By comparing (4) and (6) , we get

$y = 2x$

Until this step , equations are already easy to solve

So we can get two pair of solutions which are:

$x=1 , y=z=2 , k=12$

Or

x= \frac {5}{3} , y = \frac {10}{3} , z=0 , k = \frac {100}{9)

By computing those three value,

$G(1,2,2) = 20$

$G(1.667 , 3.333 , 0 ) = 18.5185 < 20$

So ,the maximum value of $xy^2 +yz^2 +2xyz$ is

$\boxed {20}$

I am sorry for poor english , also , if my solution have any mistake please tell me , thanks !

This is how I have solved it. I would like to use AM-GM inequality on the $2xyz$ term. In particular it is true that:

• $2xyz\leq x(y^2+z^2)$ and the max is archived when $y=z$

• $2xyz\leq y(x^2+z^2)$ and the max is archived when $x=z$

• $2xyz\leq z(x^2+y^2)$ and the max is archived when $x=y$

But only one choice can lead to the maximum. So I rewrite the expression supposing first $y=z$ and then $x=z$ , $x=y$ .

When $y=z$ we have: $xy^2+yz^2+2xyz\leq xy^2+yz^2+x(y^2+z^2)=(5-2y)y^2+y^3+2(5-2y)y^2=15y^2-5y^3$ . So we have to maximise the single variable function $f(y)=15y^2-5y^3$ . This is straightforward: the max is equal to $20$ when $y(=z)=2$ .

The others case are similar. When $x=z$ we have that the max is $\frac{125}{54}(7\sqrt(7)-10)<20$ and when $x=y$ the max is $\frac{500}{27}<20$ . So $20$ is the max and it is archived when $z=y=2$ and when $x=5-2y=1$ .