No soy una fuente confiable, pero creo que puedo demostrar el siguiente teorema: $$\sum_{cyc}\frac{a_{1}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}+k_{2}}$$
DEMOSTRACIÓN
Primero necesitaremos lo siguiente
Lema 1.
$$\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\sum_{cyc}a_{1}a_{2}^{\alpha}$$
Prueba.
Aplicando la desigualdad de la Reordenación en $a_{1},a_{2},...,a_{n}$ y $a_{1}^{\alpha},a_{2}^{\alpha},...,a_{n}^{\alpha}$, tenemos que $\sum_{cyc}a_{1}a_{2}^{\alpha}$ es maximizado cuando $a_{1},a_{2},...,a_{n}$ y $a_{1}^{\alpha},a_{2}^{\alpha},...,a_{n}^{\alpha}$ están ordenados de manera similar.
Por lo tanto, podemos afirmar que
$$\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\sum_{cyc}a_{1}a_{2}^{\alpha}$$
Por otro lado, necesitamos lo siguiente
Lema 2.
$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
Prueba.
Establecemos que
$$\frac{a_{1}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}+\frac{n}{m}=\frac{a_{1}}{k_{1}}$$
Operando, obtenemos que
$$\frac{a_{1}^{\alpha+1}m+n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)}{m\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)}=\frac{a_{1}}{k_{1}}$$
$$k_{1}\left(a_{1}^{\alpha+1}m+n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)\right)=a_{1}m\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)$$
$$k_{1}a_{1}^{\alpha+1}m+k_{1}n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)=a_{1}mk_{1}a_{1}^{\alpha}+a_{1}mk_{2}a_{2}^{\alpha}$$
$$k_{1}n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)=a_{1}mk_{2}a_{2}^{\alpha}$$
$$\frac{n}{m}=\left(\frac{k_{2}}{k_{1}}\right)\frac{a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}$$
Por lo tanto, tenemos que
$$\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{a_{1}}{k_{1}}$$
Y posteriormente, repitiendo el proceso para cada variable, obtenemos que
$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
Ahora, estamos listos para demostrar el teorema.
Aplicando el Lema 1, derivamos que
$$\left(\frac{k_{2}}{k_{1}}\right)\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\left(\frac{k_{2}}{k_{1}}\right)\sum_{cyc}a_{1}a_{2}^{\alpha}$$
Por lo tanto, sustituyendo en la expresión de Lema 2 y operando, tenemos que
$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
$$\sum_{cyc}\frac{\left(\frac{k_{1}}{k_{1}}\right)a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
$$\sum_{cyc}\frac{\left(\frac{k_{1}+k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
$$\sum_{cyc}\frac{a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{\left(\frac{k_{1}+k_{2}}{k_{1}}\right)k_{1}}$$
$$\sum_{cyc}\frac{a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}+k_{2}}$$
Como queríamos demostrar.
El caso particular propuesto sigue de la aplicación directa del teorema.
