Problem 1608

To prove the inequality
$$\sqrt[n]{a_1a_2\cdots a_n} + \sqrt[n]{b_1b_2\cdots b_n} \le \sqrt[n]{(a_1+b_1)(a_2+b_2)\cdots (a_n+b_n)},$$
we will use the Hölder's inequality.

1. Hölder's Inequality: For positive real numbers $a_i$ and $b_i$ and for $p, q > 1$ such that $\frac{1}{p} + \frac{1}{q} = 1$, Hölder's inequality states:
$$\left( \sum_{i=1}^n a_i^p \right)^{\frac{1}{p}} \left( \sum_{i=1}^n b_i^q \right)^{\frac{1}{q}} \geq \sum_{i=1}^n a_i b_i.$$
In our case, we will use $p = q = n$.

2. Applying Hölder's Inequality: Let $a_i = x_i^n$ and $b_i = y_i^n$. Then, we have:
$$\left( \sum_{i=1}^n x_i^n \right)^{\frac{1}{n}} \left( \sum_{i=1}^n y_i^n \right)^{\frac{1}{n}} \geq \sum_{i=1}^n x_i y_i.$$
Since $a_i = x_i^n$ and $b_i = y_i^n$, we can rewrite the sums as:
$$\left( \sum_{i=1}^n a_i \right)^{\frac{1}{n}} \left( \sum_{i=1}^n b_i \right)^{\frac{1}{n}} \geq \sum_{i=1}^n (a_i^{\frac{1}{n}} b_i^{\frac{1}{n}}).$$

3. Simplifying the Expression: Notice that:
$$\left( \sum_{i=1}^n a_i \right)^{\frac{1}{n}} = \sqrt[n]{a_1 + a_2 + \cdots + a_n},$$
and similarly for $b_i$. Therefore, we have:
$$\sqrt[n]{a_1 + a_2 + \cdots + a_n} \cdot \sqrt[n]{b_1 + b_2 + \cdots + b_n} \geq \sum_{i=1}^n \sqrt[n]{a_i b_i}.$$

4. Combining the Results: We need to show that:
$$\sqrt[n]{a_1a_2\cdots a_n} + \sqrt[n]{b_1b_2\cdots b_n} \le \sqrt[n]{(a_1+b_1)(a_2+b_2)\cdots (a_n+b_n)}.$$
By the AM-GM inequality, we know that:
$$\sqrt[n]{a_1a_2\cdots a_n} \leq \frac{a_1 + a_2 + \cdots + a_n}{n},$$
and similarly for $b_i$. Therefore:
$$\sqrt[n]{a_1a_2\cdots a_n} + \sqrt[n]{b_1b_2\cdots b_n} \leq \frac{a_1 + a_2 + \cdots + a_n}{n} + \frac{b_1 + b_2 + \cdots + b_n}{n}.$$

5. Final Step: We need to show that:
$$\frac{a_1 + a_2 + \cdots + a_n}{n} + \frac{b_1 + b_2 + \cdots + b_n}{n} \leq \sqrt[n]{(a_1+b_1)(a_2+b_2)\cdots (a_n+b_n)}.$$
By the AM-GM inequality again, we have:
$$\frac{a_1 + b_1 + a_2 + b_2 + \cdots + a_n + b_n}{n} \leq \sqrt[n]{(a_1+b_1)(a_2+b_2)\cdots (a_n+b_n)}.$$

6. Equality Condition: Equality holds if and only if:
$$\frac{a_1}{b_1} = \frac{a_2}{b_2} = \cdots = \frac{a_n}{b_n}.$$

Thus, we have proven the inequality and the condition for equality.

$\blacksquare$