Is the function characterized by $f(\alpha x+(1-\alpha) y) \le f^{\alpha}(x/\alpha)f^{1-\alpha}(y)$ convex? [Edit. The question lacks certain important conditions, as kindly pointed out by NeutralElement.  Below is the amended version.  I apologize for the omissions and many thanks to NeutralElement and user254665 for helpful comments. ]
This is related to the question raised by Boby.  For a variant, consider the nonnegative function $f(x)$ satisfied by
$$
f(\alpha\, x + (1 - \alpha) \, y) \le f^{\alpha}(x/\alpha) \, f^{1-\alpha}(y),
\qquad (1)
$$
for $\alpha \in (0, 1]$ and real $x$ and $y$.
What can we say about his function?  Is it always convex?
[Edit2.  In the original question by Boby, there is one additional condition that requires that $x \ge y$.  But unfortunately I missed this condition in the previous edit.  So I'll settle for the result for the above question.  I apologize for the many omissions.  However, if anyone can comment how this condition changes the result, it would be much appreciated.  Thank you.]

Here are some observations that may or may not help.
Observation 1
With $y = x$, we have
$$
f(x) \le f(x/\alpha).  \qquad (2)
$$
This means that $f(x)$ is increasing for $x \ge 0$, and decreasing for $x \le 0$.  A corollary is that $f(0)$ is the global minimum, i.e,
$$
f(0) \le f(x). \qquad (3)
$$
Observation 2
By Young's inequality or the weighted AM-GM inequality, we have
$$
f^\alpha(x/\alpha) f^{1-\alpha}(y)
\le \alpha f(x/\alpha) + (1 - \alpha) f(y). \qquad (4)
$$
Thus, (4) and (1) require
$$
f(y + \alpha (x - y))
\le
\alpha \, f(x/\alpha) + (1 - \alpha) \, f(y).
$$
 A: Some ideas below.
Since fractional powers are defined only for non-negative numbers this problem should have in its statement $f\ge0$ which comes to $f(0)\ge0$ since $0$  is the global min.
If $f(0)=0$ then pick $y=0$ in the given condition to get $f(\alpha x)\le0$ for every $\alpha\in(0,1],\ x\in\mathbb{R}$. In this case $f\equiv0$.
It remains that $f(0)>0$. Clearly $f=k>0$ constant is a solution for this problem. 
Let us look for non-constant solutions. Again for $y=0$ we get
$$
f(\alpha x)\le f^\alpha(x/\alpha)f^{1-\alpha}(0)
$$
Assume that $f$ is non-constant. Then  $\lim_{x\to+\infty}f(x)=\sup_{x\ge0} f(x)=+\infty$. Similarly, $\lim_{x\to-\infty}f(x)=\sup_{x\le0} f(x)=+\infty$. 
More precisely, if we assume that $\lim_{x\to+\infty}f(x)=\ell=\sup_{x\ge0} f(x)<+\infty$ then $\ell\ge f(0)>0$. Pick $x=\alpha^{-2}$ and let $\alpha\to 0$ to get $\ell\le f(0)$ from which we obtain the contradiction $f\equiv f(0)$.
Now I see that a sharper inequality is obtained for $x=0$. Namely,
$$
f(\alpha x)\le f^\alpha(x)f^{1-\alpha}(0) 
$$
(I renamed $y\to x,\ 1-\alpha\to\alpha$ )
Also I see that $f(x):=e^{|x|}$ is a solution so I will be concerned in the sequel only with the convexity of $f$.
Something else. Note that $f(x)=e^{g(x)}$ is a solution of this problem iff for every $\alpha\in(0,1],\ x,y\in\mathbb{R}$, $g(\alpha x+(1-\alpha)y)\le\alpha g(x/\alpha)+(1-\alpha)g(y)$. 
This remark shows that your problem is equivalent to asking whether $g$ is convex whenever $g(\alpha x+(1-\alpha)y)\le\alpha g(x/\alpha)+(1-\alpha)g(y)$ holds for every $\alpha\in(0,1],\ x,y\in\mathbb{R}$ (and that is a simplification of your problem). 
A: Statement of the problem: What can we say about a function $f:\mathbb{R}\to\mathbb{R}$ with the property that
$$
(P)\ \ \forall \alpha\in(0,1],\ x,y\in\mathbb{R},\ x\ge y,\ \ f(\alpha x+(1-\alpha)y)\le f^{\alpha}(x/\alpha)f^{1-\alpha}(y)?
$$
Since we can take $x=y$, $f(0)$ is still the global minimum value. 
Assume that $f(0)=0$. We claim in this case that $f\equiv 0$. Indeed, take $y=0$ in (P) to get $\forall \alpha\in(0,1],\ x\ge 0,\ \ f(\alpha x)\le 0$ from which $f(x)=0$ for every $x\ge0$. Similarly, make $x=0$  in (P) to get $\forall \alpha\in(0,1],\ y\le 0,\ \ f((1-\alpha) y)\le 0$ from which $f(y)=0$ for every $y\le0$.
It remains that $f(0)>0$. Clearly $f=k>0$ constant is a solution for this problem.
We claim that if $f$ is non-constant on $\mathbb{R}_{+}$ then $\lim_{x\to+\infty}f(x)=+\infty$.
Again for $y=0$ we get
$$
\forall \alpha\in(0,1],\ x\ge 0,\ \ f(\alpha x)\le f^\alpha(x/\alpha)f^{1-\alpha}(0)
$$
If we assume that $\lim_{x\to+\infty}f(x)=\ell=\sup_{x\ge0} f(x)<+\infty$ then $\ell\ge f(0)>0$. Pick $x=\alpha^{-2}$ to get 
$$
\forall \alpha\in(0,1],\ \ f(\alpha^{-1})\le f^\alpha(\alpha^{-3})f^{1-\alpha}(0)\le\ell^{\alpha} f^{1-\alpha}(0)
$$
and pass to supremum over $\alpha$ to get $\ell\le f(0)$ from which we get $f\equiv f(0)$.
Similarly, if $f$ is non-constant on $\mathbb{R}_{-}$ then $\lim_{x\to-\infty}f(x)=+\infty$. (This time pick $x=0$.)
However, let us apply the natural logarithm to the inequality to get for $g(x)=\ln f(x)$ that
$$
(PL)\ \ \ \forall \alpha\in(0,1],\ x,y\in\mathbb{R},\ x\ge y,\ \ g(\alpha x+(1-\alpha)y)\le {\alpha}g(x/\alpha)+(1-\alpha)g(y)
$$
We have $f$ has (P) iff $g$ has (PL); both $f$  and $g$ satisfy (PL); and we are looking for non-constant solutions. 
Again from $x=y$ we know that $g(0)$ is the global min value, $g$ is decreasing on the negative $x-$axis and increasing on the positive $x-$axis, and $\lim_{x\to\pm\infty}g(x)=+\infty$. We can assume that $g$ is non-negative with $g(0)=0$ otherwise we replace it by $g-g(0)$. 
Because $0$ is a global minimum we have:
$g$ is convex on $\mathbb{R}$ iff $g$ is convex on $\mathbb{R}_{-}$ and $g$ is convex on $\mathbb{R}_{+}$.
Consider the function $g(x)=\sqrt{x}$, for $x\ge0$, $g(x)=0$, for
$x<0$. 
Then for $\alpha\in(0,1)$, $x\ge y$ we consider the cases: 
(A) $\alpha x+(1-\alpha)y\le0$,
(B) $\alpha x+(1-\alpha)y>0$, $x\ge0\ge y$ and 
(C) $\alpha x+(1-\alpha)y>0$, $x\ge y\ge0$.
Condition (PL) is easily verified in case (A) because $g(\alpha x+(1-\alpha)y)=0$ and $g\ge0$. 
In case (B) we have $g(\alpha x+(1-\alpha)y)\le\alpha g(x/\alpha)+(1-\alpha)g(y)\Leftrightarrow\sqrt{\alpha x+(1-\alpha)y}\le\alpha\sqrt{x/\alpha}=\sqrt{\alpha x}$
 which is true because $\alpha x+(1-\alpha)y>0$ and $x\ge0\ge y$. 
In case (C) 
$$
g(\alpha x+(1-\alpha)y)\le\alpha g(x/\alpha)+(1-\alpha)g(y)\Leftrightarrow
$$
$$
\sqrt{\alpha x+(1-\alpha)y}\le\alpha\sqrt{x/\alpha}+(1-\alpha)\sqrt{y}=\sqrt{\alpha x}+(1-\alpha)\sqrt{y}\Leftrightarrow
$$
$$
\alpha x+(1-\alpha)y\le\alpha x+(1-\alpha)^{2}y+2(1-\alpha)\sqrt{\alpha xy}\Leftrightarrow\alpha y\le2\sqrt{\alpha xy}
$$
which is true because $\alpha\le\sqrt{\alpha}$ and $y\le\sqrt{xy}$.
Hence $g$ satisfies (PL) and is not convex (on $\mathbb{R}$). Therefore
$f(x)=e^{\sqrt{x}}$, for $x\ge0$, $f(x)=1,$ for $x<0$ is continuos,
satisfies (P) but is not convex (it is concave on $[0,1]$). So your
modified question has a negative answer.
