Question

Prove the following inequality:

Answer 1

\[\sqrt{x_{1}+x_{2}+...+x_{n}} \le \sqrt{x_{1}} + \sqrt{x_{2}} +...+ \sqrt{x_{n}}\]

Answer 2

\[\forall x_{1}, x_{2}, ..., x_{n} > 0, n \in \mathbb{N}, n \ge 2\]

Answer 3

Okay, I'm dizzy :(

Answer 4

I need to rethink this... @ParthKohli ?

Answer 5

Okay, I was wrong.

Answer 6

Yes...

Answer 7

Why didn't you say so? :(

Answer 8

i mean point it out save me some dizzy-time ^.^

Answer 9

@mukushla AM-GM? ;-)

Answer 10

It took you that long to type that? :/

Answer 11

:-P

Answer 12

okay, enough of this...
take it from me, for all positive numbers p and q
\[\huge p^2 > q^2 \iff p > q\]

Answer 13

\[\large \left(\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}...+\sqrt{x_n}\right)^2=x_1+x_2+x_3...+x_n+f(x)\]\[f(x) > 0\]

Answer 14

So...
\[\large x_1+x_2+x_3...+x_n \le x_1+x_2+x_3+...x_n+f(x)\]Getting the square root of both sides would keep the inequality the same
\[\large \sqrt{x_1+x_2+x_3...+x_n}\le\sqrt{x_1}+\sqrt{x_2}+\sqrt{x_3}...+\sqrt{x_n}\]

Answer 15

^nice. My first idea was to assume x1 = x2 = ... xn so just call them all x

Then \[x+x+x+... = n x\] and \[\sqrt x+\sqrt x +... = n \sqrt x\]

so \[\sqrt{x_{1}+x_{2}+...+x_{n}} \le \sqrt{x_{1}} + \sqrt{x_{2}} +...+ \sqrt{x_{n}} \] becomes \[\large \sqrt {nx} \le n \sqrt x\]