Answer to Question #114739 in Algebra for ANJU JAYACHANDRAN

Question

Prove that ((x+y+z)/3)^(x+y+z) ≤ x^x y^y z^z ≤ ((x^2+y^2+z^2)/(x+y+z))^(x+y+z)

Answer 1

Due to the domain of the function $a^x,$ namely $a > 0$, we obtain that $x,y,z>0$.

Let's prove the first inequality:

$(\frac{x+y+z}{3})^{(x+y+z)}\le x^xy^yz^z$. By dividing both parts by $A^{x+y+z} = (\frac{x+y+z}{3})^{(x+y+z)}$, we obtain the following:

$(\frac{x}{A})^{x}(\frac{y}{A})^{y}(\frac{z}{A})^{z} \ge 1.$ Since $A>0$, we can write the following:

$(\frac{x}{A})^{\frac{x}{A}}(\frac{y}{A})^{\frac{y}{A}}(\frac{z}{A})^{\frac{z}{A}} \ge 1.$ Let's consider the following function: $u(t,u,v) = t^tu^uv^v$, but where $t+u+v = 3$ , respectively to our case. Let's find its local minima and maxima of this function(to prove it is $\ge 1, t,u,v \in R$ ). Using the method of Lagrange multipliers we obtain the following system:

$\begin{cases} t^tu^uv^v(\ln{t} + 1) - \lambda = 0 \\ t^tu^uv^v(\ln{u} + 1) - \lambda = 0 \\ t^tu^uv^v(\ln{v} +1) - \lambda = 0 \\ t+u+v=3 \end{cases}$ . It is obviously inferred from the first 3 equations, that $t = u =v$. Using the 4th equation, we obtain that $t = u = v = 1$ . Let's figure out if this is a local maximum or minimum. Using the Hessian matrix method we obtain that it is local minimum. But, the inequality is true at $(1,1,1)$. Since the function had the minimum at $(1,1,1)$, we obtain that initial inequality is true for every $(t, u, v)$, and, hence, for each $(x,y,z) \in R^3$.

Let's now prove the 2nd inequality:

$x^xy^yz^z \le (\frac{(x^2 + y^2 + z^2)}{x+y+z})^{x+y+z}$ . Using the Weighted AM-GM Inequality, for $\omega_1 = x, \omega_2 = y, \omega_3 = z,$ we obtain: $(x^xy^yz^z)^{\frac{1}{x+y+z}} \le \frac{x\cdot x + y\cdot y + z \cdot z}{x+y+z} = \frac{x^2 + y^2 + z^2}{x+y+z}$ . Since $x+y+z > 0$ , raising both parts of the inequality to the $x+y+z$ power, we obtain the following:

$x^xy^yz^z \le (\frac{x^2 + y^2 +z^2}{x+y+z})^{x+y+z}$ .