A better bound, namely $\ \ \displaystyle a+b\le\frac{2\sqrt n}{\sqrt n+1}\ \ $ is obtained as follows.

The cubes of edge lengths $a$ and $b$ contain inscribed balls of diameters $a$ and $b$, respectively. Now, forget these two cubes for a while and maximize the sum $a+b$ of the balls' diameters, assuming that the (variable) balls are contained in the unit cube and their interiors are disjoint. Obviously, the maximum possible value of $a+b$ will be an upper bound for the sum of the edges of the cubes. Easy to see, in the optimal configuration of the balls

(1) the two balls must be tangent to each other,

(2) the balls' centers must lie on the main diagonal of the unit cube,

and

(3) each of the two balls must be tangent to the unit cube's boundary.

A brief explanation for (2): the center of the first ball lies somewhere in a cube of edge length $1-a$, concentric with, and parallel to the unit cube. Likewise, the other ball's center is confined to a cube of edge length $1-b$, concentric with, and parallel to the unit cube. The maximum distance between the balls' centers is reached when the centers lie in the "opposite" corners of their confining cubes.

In this position, by (1), the distance between the balls' centers must be $a+b$. Easy to calculate, $a+b=\frac{2\sqrt n}{\sqrt n+1}$.

Somewhat surprisingly, there is a continuum of optimal configurations, where $\frac{\sqrt n-1}{\sqrt n+1}\le a\le1$, $b=\frac{2\sqrt n}{\sqrt n+1}-a$.

Neither this, nor the bound given by asatzhh (above) proves the conjectured inequality for cubes, since each of them gives values greater than $1$ for every $n\ge2$, and they both monotonically approach $2$ from below as $n\to\infty$. However, $$\frac{2\sqrt n}{\sqrt n+1}<2^{\frac{n-1}{n}}\ {\rm for\ every}\ n\ge2.$$ For $n=3$, $\frac{2\sqrt n}{\sqrt n+1}\approx 1.26$, while $2^{\frac{n-1}{n}}\approx 1.58$.

and for $n=9$, $\frac{2\sqrt n}{\sqrt n+1}=1.5$ while $2^{\frac{n-1}{n}}\approx 1.85$.

A better bound, namely $\ \ \displaystyle a+b\le\frac{2\sqrt n}{\sqrt n+1}\ \ $ is obtained as follows.

The cubes of edge lengths $a$ and $b$ contain inscribed balls of diameters $a$ and $b$, respectively. Now, forget these two cubes for a while and maximize the sum $a+b$ of the balls' diameters, assuming that the (variable) balls are contained in the unit cube and their interiors are disjoint. Obviously, the maximum possible value of $a+b$ will be an upper bound for the sum of the edges of the cubes. Easy to see, in the optimal configuration of the balls

(1) the two balls must be tangent to each other,

(2) the balls' centers must lie on the main diagonal of the unit cube,

and

(3) each of the two balls must be tangent to the unit cube's boundary.

A brief explanation for (2): the center of the first ball lies somewhere in a cube of edge length $1-a$, concentric with, and parallel to the unit cube. Likewise, the other ball's center is confined to a cube of edge length $1-b$, concentric with, and parallel to the unit cube. The maximum distance between the balls' centers is reached when the centers lie in the "opposite" corners of their confining cubes.

In this position, by (1), the distance between the balls' centers must be $a+b$. Easy to calculate, $a+b=\frac{2\sqrt n}{\sqrt n+1}$.

Somewhat surprisingly, there is a continuum of optimal configurations, where $\ \frac{\sqrt n-1}{\sqrt n+1}\le a\le1$, $\ b=\frac{2\sqrt n}{\sqrt n+1}-a$.

Neither this, nor the bound given by asatzhh (above) proves the conjectured inequality for cubes, since each of them gives values greater than $1$ for every $n\ge2$, and they both approach $2$ monotonically from below as $n\to\infty$. However, $$\frac{2\sqrt n}{\sqrt n+1}<2^{\frac{n-1}{n}}\ \ {\rm for\ every}\ \ n\ge2.$$ For $n=3,\ $ $\frac{2\sqrt n}{\sqrt n+1}\approx 1.26,\ $ while $\ 2^{\frac{n-1}{n}}\approx 1.58$,

and for $n=9,\ $ $\frac{2\sqrt n}{\sqrt n+1}=1.5,\ $ while $\ 2^{\frac{n-1}{n}}\approx 1.85$.

A better bound, namely $\ \ \displaystyle a+b\le\frac{2\sqrt n}{\sqrt n+1}\ \ $ is obtained as follows.

The cubes of edge lengths $a$ and $b$ contain inscribed balls of diameters $a$ and $b$, respectively. Now, forget these two cubes for a while and maximize the sum $a+b$ of the balls' diameters, assuming that the (variable) balls are contained in the unit cube and their interiors are disjoint. Obviously, the maximum possible value of $a+b$ will be an upper bound for the sum of the edges of the cubes. Easy to see, in the optimal configuration of the balls

(1) the two balls must be tangent to each other,

(2) the balls' centers must lie on the main diagonal of the unit cube,

and

(3) each of the two balls must be tangent to the unit cube's boundary.

A brief explanation for (2): the center of the first ball lies somewhere in a cube of edge length $1-a$, concentric with, and parallel to the unit cube. Likewise, the other ball's center is confined to a cube of edge length $1-b$, concentric with, and parallel to the unit cube. The maximum distance between the balls' centers is reached when the centers lie in the "opposite" corners of their confining cubes.

In this position, by (1), the distance between the balls' centers must be $a+b$. Easy to calculate, $a+b=\frac{2\sqrt n}{\sqrt n+1}$.

Somewhat surprisingly, there is a continuum of optimal configurations, where $\frac{\sqrt n-1}{\sqrt n+1}\le a\le\frac{2\sqrt n}{\sqrt n+1}$, $b=\frac{2\sqrt n}{\sqrt n+1}-a$.

Neither this, nor the bound given by asatzhh (above) proves the conjectured inequality for cubes, since each of them gives values greater than $1$ for every $n\ge2$, and they both monotonically approach $2$ from below as $n\to\infty$. However, $$\frac{2\sqrt n}{\sqrt n+1}<2^{\frac{n-1}{n}}\ {\rm for\ every}\ n\ge2.$$ For $n=3$, $\frac{2\sqrt n}{\sqrt n+1}\approx 1.26\ldots$, while $2^{\frac{n-1}{n}}\approx 1.58\ldots$

and for $n=9$ $\frac{2\sqrt n}{\sqrt n+1}=1.5$ while $2^{\frac{n-1}{n}}\approx 1.85\ldots$.

A better bound, namely $\ \ \displaystyle a+b\le\frac{2\sqrt n}{\sqrt n+1}\ \ $ is obtained as follows.

The cubes of edge lengths $a$ and $b$ contain inscribed balls of diameters $a$ and $b$, respectively. Now, forget these two cubes for a while and maximize the sum $a+b$ of the balls' diameters, assuming that the (variable) balls are contained in the unit cube and their interiors are disjoint. Obviously, the maximum possible value of $a+b$ will be an upper bound for the sum of the edges of the cubes. Easy to see, in the optimal configuration of the balls

(1) the two balls must be tangent to each other,

(2) the balls' centers must lie on the main diagonal of the unit cube,

and

(3) each of the two balls must be tangent to the unit cube's boundary.

A brief explanation for (2): the center of the first ball lies somewhere in a cube of edge length $1-a$, concentric with, and parallel to the unit cube. Likewise, the other ball's center is confined to a cube of edge length $1-b$, concentric with, and parallel to the unit cube. The maximum distance between the balls' centers is reached when the centers lie in the "opposite" corners of their confining cubes.

In this position, by (1), the distance between the balls' centers must be $a+b$. Easy to calculate, $a+b=\frac{2\sqrt n}{\sqrt n+1}$.

Somewhat surprisingly, there is a continuum of optimal configurations, where $\frac{\sqrt n-1}{\sqrt n+1}\le a\le1$, $b=\frac{2\sqrt n}{\sqrt n+1}-a$.

Neither this, nor the bound given by asatzhh (above) proves the conjectured inequality for cubes, since each of them gives values greater than $1$ for every $n\ge2$, and they both monotonically approach $2$ from below as $n\to\infty$. However, $$\frac{2\sqrt n}{\sqrt n+1}<2^{\frac{n-1}{n}}\ {\rm for\ every}\ n\ge2.$$ For $n=3$, $\frac{2\sqrt n}{\sqrt n+1}\approx 1.26$, while $2^{\frac{n-1}{n}}\approx 1.58$.

and for $n=9$, $\frac{2\sqrt n}{\sqrt n+1}=1.5$ while $2^{\frac{n-1}{n}}\approx 1.85$.