Let $a,b\geq 0$ and let $(p,q)$ be conjugate exponents, which means $p,q>0$ and $p^{-1}+q^{-1}=1$. Then, we have the so-called Young's inequality:
$$ab\le \frac{a^{p}}{p}+\frac{b^{q}}{q}$$
Now, if $n\in\mathbb{Z}^{+}$, we deduce easily from the binomial theorem that the following holds:
$$(a+b)^{n}\le C_{n}(a^{n}+b^{n}) \tag{1}$$
where $C_{n}>0$ is a positive constant only depending on $n$.
My goal is to prove $(1)$ for real $n\geq 0$. Let's denote $r=n$ to remember $r$ is real and not especially an integer. We consider only $r\in\mathbb{R}^{+}\setminus\mathbb{Z}^{+}$. Let $\gamma=\lfloor{r}\rfloor\in\mathbb{Z}$. Obviously, there exists $\alpha\in (0,1)$ such that $r=\alpha \gamma + (1-\alpha)(\gamma +1)$.
Now, as $\left(\frac{1}{\alpha}\right)^{-1}+\left(\frac{1}{1-\alpha}\right)^{-1}=1$, we can apply Young's inequality, which leads to:
\begin{align*} (a+b)^{r}&=(a+b)^{\alpha \gamma + (1-\alpha)(\gamma +1)}\\ &\le \frac{(a+b)^{\frac{1}{\alpha}\alpha \gamma}}{1/\alpha} + \frac{(a+b)^{\frac{1}{1-\alpha}(1-\alpha)(\gamma +1)}}{1/(1-\alpha)}\\ &=\alpha (a+b)^{\gamma} + (1-\alpha)(a+b)^{\gamma +1}\\ &\le \alpha C_{\gamma}(a^{\gamma}+b^{\gamma})+(1-\alpha)C_{\gamma+1}(a^{\gamma+1}+b^{\gamma+1}) \end{align*}
However, I am stuck here. How to conclude if this is even a good approach? Any help is appreciated. If possible, I would like to avoid the so-called "generalized binomial theorem".
Not sure if this is what you are looking for, but this is a method to prove (1). Notice by Young's inequality \begin{equation} x^p y^q < \frac{p}{p+q}x^{p+q} + \frac{q}{p+q} y^{p+q}. \end{equation} If this isn't clear, consider $a = x^p$ and $b = y^p$ in the formulation you gave. Your conjugate exponents in this case would be $\frac{p+q}{p}$ and $\frac{p+q}{q}$. Since every term in the expansion of $(a+b)^n$ is of the form $a^p b^q$ with $p+q=n$, every term in the expansion may be bounded by a term of the form $\frac{p}{p+q} a^n + \frac{q}{p+q}b^n$. Since $p,q$ are dependent only on $n$, the result follows. Hope that helps.