You could prove it along this lines:
Suppose, $x \ne y$ are both feasible and $w \cdot x = w \cdot y$. Then, since the feasible set is strictly convex, $x/2 + y/2$ lies in the interior of the feasible set. Set $z = x/2 + y/2 - \varepsilon \, w$ for small $\varepsilon > 0$. This shows $z$ is feasible and $w \cdot z < w \cdot x$.
Hence, the minimizer is unique.
hmmmmm. I am not sure. The property you mention is not generally a property (in general) of either concave or convex functions, but actually of sub-additive functions. However, I am inclined to believe that the conjecture, or the book may be correct for the following reasoning.
Technically, (on the average, to abuse terminology) a property that would most likely be shared by concave functions over the positive reals, rather then convex functions. Unless its one of these sub-linear or super-linear functions which are some admixture of both (and almost, both,'convex and concave', or super-additive, and sub-additive) over $[0,1]$)
Notice that in $(1)$ the scope of the multiplicative elements $a$ and $(1-a)$ is within $F$ not outside $F$ as in $(2)$ and $(3)$. If one can show that $F$ is sub-additive or that F(tx)>=tF(x) it should be.
$(1)f(ax+(1−a)y)≤f(ax)+f((1−a)y)$
Convex: $f(ax+(1−a)y) ≤a f(x)+(1−a)f(y)$
Concave: $f(ax+(1−a)y)\geq af(x)+(1−a)f(y)$
I agree that $(1)$ definitely Seems wrong for concave $F$ but it actually may not be. Remember that the "above" $(1)$ is a property that function, $F$ 'sub-additive over the positive reals' may have.
Concave functions are sub-additive, over the positive reals, when $f(0) ≥ 0$,see https://en.wikipedia.org/wiki/Concave_function. See pt 10.
Although the domain only specifies positive- reals, not-non-negative reals and may not include the $0$. That is the only the issue. It also not a closed domain and range either.
However, given the positive domain, and strict monotonic increasing-ness, it might have the same effect, and the conjecture MIGHT (I stress) be correct. It will also be strictly quasi convex and strictly quasi concave.
If $F$ is concave and,$f(0) ≥ 0$ then $F$ is sub-additive over the positive reals; so the book, is "arguably" correct.
I (stress) arguably, as the the domain, of the function, only specifies positive- reals, not-negative reals. ie including $0$. Maybe strict mono-tonicity, and posi-tivity may help in this case. Notice, that given sub-additivity.
Best Answer
It is useful for this problem that the (strict) concavity of a function $\mathbb{R}^n \to \mathbb{R}$ is equivalent to its (strict) concavity on every one-dimensional affine subspace---that is, on every line.
Now, $-y^2 - z^2$ is strictly concave on every line not parallel to the line generated by $(1, 0, 0)$, and so by the proposition $f(x, y, z)$ is strictly concave on such lines. Also, $-e^{x + y + z}$ is strictly concave on every line not parallel to the plane $x + y + z = 0$, so again by the proposition $f(x, y, z)$ is strictly concave here. Because the line generated by $(1, 0, 0)$ and the plane defined by $x + y + z = 0$ together generate $\mathbb{R}^3$, every line falls into one of these two classes, so this argument is enough.