Minimum of function of $3$ variables

Question

If $xyz = a^3$ then show that the minimum value of $x^2+y^2+z^2$ is $3a^2$.

I have tried this problem using the identity $(x + y + z)^2$ but I am not satisfied with my approach. Any other method of solving it.

Answer 1

Using AM, GM inequality, $$\dfrac{x^2+y^2+z^2}3\ge\sqrt[3]{x^2y^2z^2}$$

Answer 2

Using a Lagrange multiplier,

$$\mathcal{L} (x,y,z,\lambda) := \frac{1}{2} (x^2 + y^2 + z^2) + \lambda (x y z - a^3)$$

Taking the partial derivative and finding where they vanish,

$$\begin{array}{rl} x + \lambda y z &= 0\\ y + \lambda x z &= 0\\ z + \lambda x y &= 0\\ x y z &= a^3\end{array}$$

Multiplying the first equation by $x$, the second by $y$ and the third by $z$,

$$\begin{array}{rl} x^2 + \lambda x y z &= 0\\ y^2 + \lambda x y z &= 0\\ z^2 + \lambda x y z &= 0\\ x y z &= a^3\end{array}$$

After some work, we conclude that the minimum is attained at

$$x^2 = y^2 = z^2 = a^2$$

and, thus, the minimum is $3 a^2$.

Answer 3

Or use "Lagrange Multipliers". We want to minimize $f(x,y,z)= x^2+ y^2+ z^2$ subject to the constraint $g(x,y,z)= xyz= a^3$. First find the gradient of each function: $\nabla f= 2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}$ and $\nabla g= yz\vec{i}+ xz\vec{j}+ xy\vec{k}$. It can be shown, geometrically, that where f is a min or max, subject to constraint g constant, those two vectors are parallel- that is, one is a multiple of the other- $\nabla f= \lambda\nabla g$ so $2x\vec{i}+ 2y\vec{j}+ 2z\vec{k}= \lambda(yz\vec{i}+ xz\vec{j}+ xy\vec{k})= \lambda\nabla g$ for some constant, $\lambda$, the "Lagrange multiplier".

From that, $2x= \lambda yz$, $2y= \lambda xz$, and $2z= \lambda xy$. Since a specific value for $\lambda$ is not necessary for a solution, often it is simplest to first eliminate $\lambda$ by dividing one equation by another. Dividing the first equation by the second, $\frac{x}{y}= \frac{y}{x}$. Dividing the third equation by the second, $\frac{y}{z}= \frac{z}{y}$. Those give $x^2= y^2= z^2$ so that each of y and z is $\pm$ x.

From $xyz= a^3$ we have either $x^3= a^3$ or $-x^3= a^3$. Taking only the real roots, either x= a or x= -a. We have solutions: (a, a, a), (a, -a, -a), (-a, -a, a), or (-a, a, -a).

In any case, the minimum value of $x^2+ y^2+ z^2$ is $3a^2$, NOT $a^3$.