Prove that $\sqrt [a_1] \alpha+\sqrt [a_2] \alpha+ \dots +\sqrt [a_n] \alpha \le \alpha +n-1.$

Question

Let $\alpha$ be a positive number and $a_1,a_2, \dots, a_n$ positive non-zero integers such that $\frac {1}{a_1}+ \frac {1}{a_2} + \dots+ \frac {1}{a_n}=1$. Prove that $\sqrt [a_1] \alpha+\sqrt [a_2] \alpha+ \dots +\sqrt [a_n] \alpha \le \alpha +n-1.$

Answer 1

Note that if $n$ is a natural number, then $(1+x)^{1/n}\leq 1+x/n$. Thus, letting $\beta = \alpha-1$, $$\alpha^{1/\alpha_1}+...+\alpha^{1/\alpha_n} = (1+\beta)^{1/\alpha_1}+...+(1+\beta)^{1/\alpha_n}$$ $$\leq \left(1+\frac{\beta}{\alpha_1}\right)+...+\left(1+\frac{\beta}{\alpha_n}\right) = n+\beta\left(\frac{1}{\alpha_1}+...+\frac{1}{\alpha_n}\right)$$ $$=n+\beta = n+\alpha-1$$

Answer 2

By Bernoulli (https://en.wikipedia.org/wiki/Bernoulli%27s_inequality) we obtain: $$\sum_{i=1}^n\alpha^{\frac{1}{a_i}}=\sum_{i=1}^n(1+\alpha-1)^{\frac{1}{a_i}}\leq\sum_{i=1}^n\left(1+\frac{1}{a_i}(\alpha-1)\right)=n+\alpha-1.$$