Prove that $ a^{\log_b(a)} + b^{\log_c(b)} + c^{\log_a(c)} \geq a^{\sqrt{\log_b(a)}} + b^{\sqrt{\log_c(b)}} + c^{\sqrt{\log_a(c)}} \geq a + b + c$

Question

Let $ a, b, c > 1 $. Prove that $ a^{\log_b(a)} + b^{\log_c(b)} + c^{\log_a(c)} \geq a^{\sqrt{\log_b(a)}} + b^{\sqrt{\log_c(b)}} + c^{\sqrt{\log_a(c)}} \geq a + b + c. $

My approach: I denoted $ \log_b(a) = x , \log_c(b) = y , and \log_a(c) = z$. Then our inequality becomes $ a^x + b^y + c^z \geq a^{\sqrt{x}} + b^{\sqrt{y}} + c^{\sqrt{z}} \geq b^x + c^y + a^z. $ I tried applying rearrangement inequality, but I couldn't find the ordering of $(x, y, z)$.

Answer 1

For the left inequality:

Let $x := \frac12 \ln \ln a, y := \frac12 \ln \ln b, z := \frac12 \ln \ln c$. The desired left inequality is written as $$\mathrm{e}^{\mathrm{e}^{4x - 2y}} + \mathrm{e}^{\mathrm{e}^{4y - 2z}} + \mathrm{e}^{\mathrm{e}^{4z - 2x}} \ge \mathrm{e}^{\mathrm{e}^{3x - y}} + \mathrm{e}^{\mathrm{e}^{3y - z}} + \mathrm{e}^{\mathrm{e}^{3z - x}}. \tag{1}$$

Note that $u \mapsto \mathrm{e}^{\mathrm{e}^{u}}$ is convex on $\mathbb{R}$ (easy by checking second derivative).

Let $U := 4x - 2y, V := 4y - 2z, W := 4z - 2x, X := 3x - y, Y := 3y - z, Z := 3z - x$.

Since (1) is cyclic, assume that $x = \max(x, y, z)$.

We split into two cases.

Case 1. $x \ge y \ge z$

Clearly, we have $U + V + W = X + Y + Z$.

We have $U - W = 6x - 2y - 4z \ge 0$, $V - W = 4y - 6z + 2x \ge 0$, $X - Z = 4x - y - 3z \ge 0$, $Y - Z = 3y - 4z + x \ge 0$ which results in $W = \min(U, V, W)$ and $Z = \min(X, Y, Z)$. We have $U + V - (X + Y) = x - z \ge 0$.

Also, we have $U - X = x - y \ge 0$ and $V - Y = y - z \ge 0$ which results in $\max(U, V) \ge \max(X, Y)$.

Thus, $(X, Y, Z)$ is majorized by $(U, V, W)$. By Karamata's inequality, (1) is true.

Case 2. $x \ge z \ge y$

Clearly, $U + V + W = X + Y + Z$.

We have $U - W = 6x - 2y - 4z \ge 0$, $U - V = 4x - 6y + 2z \ge 0$, $X - Z = 4x - y - 3z \ge 0$, $X - Y = 3x - 4y + z \ge 0$ which results in $U = \max(U, V, W)$ and $X = \max(X, Y, Z)$. We have $U - X = x - y \ge 0$.

Also, we have $U + V - (X + Y) = x -z \ge 0$, and $U + W - (X + Z) = z - y \ge 0$ which results in $\max(U + V, U + W) \ge \max(X + Y, X + Z)$.

Thus, $(X, Y, Z)$ is majorized by $(U, V, W)$. By Karamata's inequality, (1) is true.

We are done.