Minimizing An Outlandish Expression

Relevant wiki: Jensen's Inequality

Well let me answer in a different way than the title expects me to,

Consider $\displaystyle g(x)=x\ln x$ and since $x>0$ we have $g''(x)>0$ so $g$ is convex and by Jensen's inequality,

$\displaystyle g(x)+g(y)+g(z)\ge 3g(\frac{x+y+z}{3})$ which implies $\displaystyle f(x,y,z) = x \ln x + y \ln y + z \ln z \ge 3e$ with equality occuring at $x=y=z=e$ .

Alternative approach is more interesting since we need to prove a minimum exists and investigate whether a global or local minima it is.

Due to $\{(x,y,z) \in \mathbb{R}^3; \space x >0, y >0 , z >0\}$ is an open set, a global minimum is a local minimum. We are going to use Lagrange multipliers and hessian matrix . Let $L(x,y,z) = x\ln(x) + y\ln(y) + z\ln(z) + \lambda (x + y + z - 3a)$ with $a > 0$ constant. $\dfrac{d}{dx} L(x,y,z) = \ln(x) + 1 + \lambda = 0,$ $\dfrac{d}{dy} L(x,y,z) = \ln(y) + 1 + \lambda = 0,$ $\dfrac{d}{dz} L(x,y,z) = \ln(z) + 1 + \lambda = 0,$ Clearing $\lambda$ we get $x = y = z$ . Because of, $x + y + z = 3a$ we can infere that $x = y = z = a$ . To see that $(a,a,a)$ is a minimum for $f$ we'll use the hessian matrix $d^2 \space L(a,a,a) = \begin{pmatrix} \frac{1}{a} & 0 & 0 \\ 0 & \frac{1}{a} & 0 \\ 0 & 0 & \frac{1}{a} \end{pmatrix}$ which is a defined positive matrix, and therefore $f$ attains a global minimum at $(a,a,a)$ . In this case, we have a minimum at $(e,e,e)$ and the minimum is $f(e,e,e) = 3e\ln(e) = 3e$ .

Note.- $d^2 \space L(a,a,a) = \begin{pmatrix} \frac{1}{a} & 0 & 0 \\ 0 & \frac{1}{a} & 0 \\ 0 & 0 & \frac{1}{a} \end{pmatrix}$ is a defined positive matrix means $d^2 \space L(a,a,a) (h,h)= (h_1, h_2 , h_3) \cdot \begin{pmatrix} \frac{1}{a} & 0 & 0 \\ 0 & \frac{1}{a} & 0 \\ 0 & 0 & \frac{1}{a} \end{pmatrix} \cdot \left(\begin{array} {c c c} h_1 \\ h_2 \\ h_3 \end{array}\right) = \frac{1}{a} h_1^2 + \frac{1}{a} h_2^2 + \frac{1}{a} h_3^2 > 0$ for all $h = (h_1, h_2 , h_3) \neq 0$

Since $x, y, z >0$ , we can apply AM-GM inequality as follows.

$\begin{aligned} x \ln x + y \ln y + z \ln z & \ge 3 \sqrt[3]{ x \ln x \cdot y \ln y \cdot z \ln z } & \small \color{#3D99F6}{\text{Equality occurs when }x=y=z=e} \\ & \ge 3e \approx \boxed{8.15485} \end{aligned}$

Here is a simple elementary proof. We have, $x \ln \frac{e}{x} + y\ln \frac{e}{y} + z \ln \frac{e}{z}\stackrel{(a)}{\leq} x (\frac{e}{x}-1) + y(\frac{e}{y}-1) + z (\frac{e}{z}-1)\stackrel{(b)}{=}0,$ where in (a) we have used the fact that $\ln(x) \leq x-1, \forall x\geq 0$ and in (b) we have used the constraint that $x+y+z=3e$ . Rearranging the above, we have $x \ln(x) + y \ln(y) + z \ln(z) \geq x+y+z=3e$ It is easy to see that the bound is achieved when $x=y=z=e. \hspace{50pt} \blacksquare$