Question on Lagrange multipliers method

Question

My understanding
Finding the minimum of the function using Lagrange multipliers method

\begin{align}
f(x,y,z) &= (y^2+z^2-x^2)^2+4x^2(x+y)^2 \\
x+y+z &= 1 \\
x &\leq y \\
x, y, z &\geq 0
\end{align}

My approach is to let

$$h(x,y,z)=f-\lambda g$$ where $~g=x+y+z-1=0$.

Geometrically to observe all the constraints we get a triangular shape in 1st octant in which the function $f$ has to minimize.

Now using the Lagrange method

\begin{align}
\frac{\partial h }{\partial x} &= -4x(y^2+z^2-x^2)+8x(x+y)^2+8x^2(x+y)-\lambda &=0 \\
\frac{\partial h }{\partial y} &= 4y(y^2+z^2-x^2)+8x^2(x+y)-\lambda &=0 \\
\frac{\partial h }{\partial z} &= 4z(y^2+z^2-x^2)-\lambda &=0 \\
\frac{\partial h}{\partial \lambda} &= x+y+z-1 &=0
\end{align}

Now my doubt, (1) In order to solve these above equation we first find $x,y,z$ in terms of $\lambda $ and put all those value in the last equation to find value of $\lambda, $ But
how to find $x,y,z$ in terms of $\lambda $??

(2) Can we say the given function attains its minimum in the triangle formed by all the constraints and So we first find value of f(x,y,z) on the three side of that triangle and then in order to find the internal minimum point then we apply Lagrange method?? This technique is new to me and it is given in one research paper. Any comment on this ??

he first calculate f(x,y,z) on x=0,x=y and z=0 and he apply Lagrange method to find internal minimum point, and finally he prove that minimum of f is at x=y=1/4 and z=1/2.

Any help is appreciated.
thanks

Answer 1

By setting $\lambda=4\mu$ you get $$ \left\{ \begin{array}{l} - x\left( {y^2 + z^2 - x^2 } \right) + 2x\left( {x + y} \right)^2 + 2x^2 \left( {x + y} \right) = \mu \\ y\left( {y^2 + z^2 - x^2 } \right) + 2x^2 \left( {x + y} \right) = \mu \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y + z = 1 \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ Thus, from the third equation, you have $$ \left\{ \begin{array}{l} - x\left( {y^2 + z^2 - x^2 } \right) + 2x\left( {x + y} \right)^2 + 2x^2 \left( {x + y} \right) = z\left( {y^2 + z^2 - x^2 } \right) \\ y\left( {y^2 + z^2 - x^2 } \right) + 2x^2 \left( {x + y} \right) = z\left( {y^2 + z^2 - x^2 } \right) \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y + z = 1 \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ Hence $$ \left\{ \begin{array}{l} 2x\left( {x + y} \right)^2 + 2x^2 \left( {x + y} \right) = \left( {x + z} \right)\left( {y^2 + z^2 - x^2 } \right) \\ 2x^2 \left( {x + y} \right) = \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y + z = 1 \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ From the second equation, you get $$ \left\{ \begin{array}{l} 2x\left( {x + y} \right)^2 + \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) = \left( {x + z} \right)\left( {y^2 + z^2 - x^2 } \right) \\ 2x^2 \left( {x + y} \right) = \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y + z = 1 \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ Therefore $$ \left\{ \begin{array}{l} 2x\left( {x + y} \right)^2 = \left( {x + y} \right)\left( {y^2 + z^2 - x^2 } \right) \\ 2x^2 \left( {x + y} \right) = \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y + z = 1 \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ Hence $$ \left\{ \begin{array}{l} \left( {x + y} \right)\left[ {2x\left( {x + y} \right) - \left( {y^2 + z^2 - x^2 } \right)} \right] = 0 \\ 2x^2 \left( {x + y} \right) = \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y + z = 1 \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$

• First case: if $x+y=0$ you easily get $z=1$ and then $y=1,x=-1$ and this not acceptable.
• Second case: $x+y\neq 0$. Then, from the first equation you have $$ \left\{ \begin{array}{l} 2x\left( {x + y} \right) = \left( {y^2 + z^2 - x^2 } \right) \\ 2x^2 \left( {x + y} \right) = \left( {z - y} \right)\left( {y^2 + z^2 - x^2 } \right) \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y + z = 1 \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ Now, by substituting the first equation in the second, you have $$ \left\{ \begin{array}{l} 2x\left( {x + y} \right) = \left( {y^2 + z^2 - x^2 } \right) \\ x^2 = x\left( {z - y} \right) \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y + z = 1 \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ thus $$ \left\{ \begin{array}{l} 2x\left( {x + y} \right) = \left( {y^2 + z^2 - x^2 } \right) \\ x\left( {x - z + y} \right) = 0 \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y + z = 1 \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ Again, from the second, with $x=0$ you get a non acceptable solution, while, if $x \neq 0$, you have $$ \left\{ \begin{array}{l} 2x\left( {x + y} \right) = \left( {y^2 + z^2 - x^2 } \right) \\ x - z + y = 0 \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y + z = 1 \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ From the second and the fourth equation, you have $$ \left\{ \begin{array}{l} 2x\left( {x + y} \right) = \left( {y^2 + z^2 - x^2 } \right) \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y = \frac{1}{2} \\ z = \frac{1}{2} \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ Now you can rewrite the first equation as $$ \left\{ \begin{array}{l} 2x\left( {x + y} \right) = \left( {y - x} \right)\left( {y + x} \right) + z^2 \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y = \frac{1}{2} \\ z = \frac{1}{2} \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ from which you get $$ \left\{ \begin{array}{l} 2x - y = \frac{1}{4} \\ z\left( {y^2 + z^2 - x^2 } \right) = \mu \\ x + y = \frac{1}{2} \\ z = \frac{1}{2} \\ x \le y \\ x,y,z \ge 0 \\ \end{array} \right. $$ And finally we have, from the first and the third $x=y=1/4$, $z=1/2$. Thus the solution of the system is what you indicate. Of course you have to check the boundary.