# Proof for $\forall x \in R^+, x^2 + y^2 + z^2 \geq xy + xz + yz$

I am doing a question to practice doing proofs with real numbers as I am still not so good at it. I ran into some problems for the following question where it is as given: $\forall x \in R^+, x^2 + y^2 + z^2 \geq xy + xz + yz$

When I tried to do it, I started off thinking it is related to $$x^2 + y^2 + z^2 - xy - xz - yz$$ and finding a way to factor the expression. However, I ran into quite a bit of trouble and I noticed that it is quite a general problem for me which is for some polynomials like this, how does one help themselves to reach the process where they realize that there is some way to factor such an expression. I had thought about a way to factor this polynomial in addition to doing various polynomials but I just couldn't find the way to express it in a way where one can use $$(x-y)^2 + (x-z)^2 + (y-z)^2$$

For instance, I thought about using $(x+y+z)^2$ and doing various forms of subtraction which was to no avail or even writing as $(x+y+z)((x-y)(\quad)+(y-z)(\quad)+(z-x)(\quad))$.

Could someone please provide some advice in terms of the approaches because it generally follows that I am not really having problems with quadratics or any polynomial with 2 variables but if I have a problem with 3 variables, I think I can easily be thrown off with questions involving more than 3 variables.

So would someone please provide some advice ? It would be greatly appreciated.

$$x^2+y^2+z^2 \ge xy+yz+zx \Leftrightarrow$$ $$\Leftrightarrow 2(x^2+y^2+z^2) \ge 2(xy+yz+zx) \Leftrightarrow$$ $$\Leftrightarrow x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+x^2 \ge 0\Leftrightarrow$$ $$\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2 \ge 0$$