How to prove an inequality using mathematical induction?

Question

I have to prove the following:

$$ \sqrt{x_1} + \sqrt{x_2} +...+\sqrt{x_n} \ge \sqrt{x_1 + x_2 + ... + x_n}$$

For every $n \ge 2$ and $x_1, x_2, ..., x_n \in \Bbb N$

Here's my attempt:

Consider $P(n): \sqrt{x_1} + \sqrt{x_2} +...+\sqrt{x_n} \ge \sqrt{x_1 + x_2 + ... + x_n}$

$$P(2): \sqrt{x_1} + \sqrt{x_2} \ge \sqrt{x_1 + x_2}$$ $$ x_1 + x_2 + 2\sqrt{x_1x_2} \ge x_1 + x_2$$ Which is true because $2\sqrt{x_1x_2} > 0$.

$$P(n + 1): \sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_n} + \sqrt{x_{n+1}} \ge \sqrt{x_1 + x_2 +...+ x_n + x_{n+1}}$$

From the hypothesis we have:

$$\sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_n} + \sqrt{x_{n+1}} \ge \sqrt{x_1 + x_2 + ... + x_n} + \sqrt{x_{n + 1}} \ge \sqrt{x_1 + x_2 +...+ x_n + x_{n+1}}$$

Squaring both sides of the right part:

$$ x_1 + x_2 + ... + x_n + x_{n + 1} + 2\sqrt{x_{n+1}(x_1 + x_2 +... + x_n)} \ge x_1 + x_2 +...+ x_n + x_{n+1} $$

Which is true, hence $P(n + 1)$ is true as well.

I'm not sure if I did it correctly?

Answer 1

I'm will just show it my way, since I find it simpler.

First, we will start off with the basis step where $n = 1$.

Basis step

$\sqrt{x_1} \geq \sqrt{x_1}$, which is of course true.

Inductive step

We let the inequality be true for any $k$, where $0 \leq k \leq n$:

$\sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_k} \geq \sqrt{x_1 + x_2 + ... + x_k}$

If this inequality must hold, the inequalty must, by induction, hold for $k + 1$:

$\sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_k} + \sqrt{x_{k + 1}} \geq \sqrt{x_1 + x_2 + ... + x_k + x_{k + 1}}$

We add $x_{k + 1}$ to both sides of the inequality:

$\sqrt{x_1} + \sqrt{x_2} + ... + \sqrt{x_k} + \sqrt{x_{k + 1}} \geq \sqrt{x_1 + x_2 + ... + x_k} + \sqrt{x_{k + 1}} \geq \sqrt{x_1 + x_2 + ... + x_k + x_{k + 1}}$

The inequality is hereby proved.

Answer 2

Without induction. For $any$ non-negative reals $x_1,...,x_n$ let $x_j=(y_j)^2$ for each $j,$ with each $y_j\ge 0.$ The inequality is then $$y_1+...+y_n\ge \sqrt {(y_1)^2+...+(y_n)^2}.$$ Since both sides are non-negative reals, this is equivalent to $$(y_1+...+y_n)^2\ge (y_1)^2+...+(y_n)^2.$$ Expand the LHS of this and you see that each term $(y_j)^2$ appears on the LHS, and (if $n>1$) all the other terms are non-negative, while $(y_1)^2,...,(y_n)^2$ are the $only$ terms on the RHS, so the LHS is $\ge$ the RHS.

E.g. For $a,b,c\ge 0$ we have $(a+b)^2=a^2+b^2+2ab\ge a^2+b^2,$ and

$(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca\ge a^2+b^2+c^2.$