Elementary proof for the inequality

Question

I have conjectured the following inequality:

$x+y+z\leq\sqrt{x^2+y^2+z^2}$ where $x,y,z\in \mathbb{R}$

I have tried to come up with an elementary proof but I failed miserably.

Can Anyone help me, please? Whether the conjecture is true or not?

Answer 1

I think the following reasoning is elementary enough.

If $x+y+z<0$ so our inequality is true.

But for $x+y+z\geq0$ it's enough to prove that $$3(x^2+y^2+z^2)\geq(x+y+z)^2$$ or $$\sum_{cyc}(2x^2-2xy)\geq0$$ or $$\sum_{cyc}(x^2-2xy+y^2)\geq0$$ or $$\sum_{cyc}(x-y)^2\geq0,$$ which is obvious.

Answer 2

Expanding on the comment, the conjecture suggested by the OP does not hold. However, for any three real number $x,y,z$, we have:

$$x+y+z\leq \sqrt{3(x^2+y^2+z^2)}$$

It can be proved using Cauchy-Schwarz:

$$(1^2+1^2+1^2)(x^2+y^2+z^2)\geq (x+y+z)^2$$

and thus:

$$\sqrt{3(x^2+y^2+z^2)}\geq |x+y+z| \geq x+y+z$$