Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.

Knowing that for any set of real numbers $x,y,z$, such that $x+y+z = 1$ the inequality $x^2+y^2+z^2 \ge \frac{1}{3}$ holds.

I spent a lot of time trying to solve this and, having consulted some books, I came to this:
$$2x^2+2y^2+2z^2 \ge 2xy + 2xz + 2yz$$ $$2xy+2yz+2xz = 1-(x^2+y^2+z^2) $$ $$2x^2+2y^2+2z^2 \ge 1 - x^2 -y^2 - z^2 $$ $$x^2+y^2+z^2 \ge \frac{1}{3}$$ But this method is very unintuitive to me and I don't think this is the best way to solve this. Any remarks and hints will be most appreciated.

Solution 1:

Cauchy- Schwarz works: $$x^2+y^2+z^2=\frac{1}{3}(1^2+1^2+1^2)(x^2+y^2+z^2)\geq\frac{1}{3}(x+y+z)^2=\frac{1}{3}$$

Solution 2:

$x^2+y^2+z^2$ only depends on the squared distance of $(x,y,z)$ from the origin and the constraint $x+y+z=1$ tells us that $(x,y,z)$ lies in a affine plane. The problem is solved by finding the distance between such plane and the origin: since the plane is orthogonal to the line $x=y=z$,

$$ \min_{x+y+z=1}x^2+y^2+z^2 = \left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^2 = \frac{1}{3}$$ and we are done.

Solution 3:

This is not a proof in itself, but if you've studied statistics, then you've seen a proof that

$$0\le V(X)=E(X^2)-E(X)^2$$

If we now consider a random variable $X$ with three equally likely values, $X=x,y$, and $z$, then we have

$$E(X)={x+y+z\over3}\qquad\text{and}\qquad E(X^2)={x^2+y^2+z^2\over3}$$

If, in addition, we assume $x+y+z=1$, then we have $E(X)={1\over3}$, which implies $E(X^2)\ge\left(1\over3\right)^2={1\over9}$, or $x^2+y^2+z^2\ge{1\over3}$.