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.

Question

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.

Answer 1

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

Answer 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.

Answer 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 $x^2+y^2+z^2\ge{1\over3}$.

Answer 4

Excellent algebraic have been given (I voted for them). Intuition can here be obtained through visual proofs.

The equation $x+y+z=1$ defines a plane. $x^2+y^2+z^2=1/3$ defines a sphere. The following visualization depicts the plane in blue, the sphere in red.

From that, you can imagine that the question could be rephrased as (in a mundane way): for any point in the plane, its distance from the $(0,0,0)$ origin is higher than $1/\sqrt{3}$? So the sphere should remain "below" the plane. Except when they meet. The symmetry of the problem tells you that the tangency point has equal coordinates $(1/\sqrt{3},1/\sqrt{3},1/\sqrt{3})$, which is one of the motivations behind the equation $(3x-1)^2+(3y-1)^2+(3z-1)^2$.

Would the sphere be bigger (a radius higher than $1/\sqrt{3}$), it would intersect the plane in more than a single tangency point.

All in all, this resorts to finding the distance of the plane to the origin, which is exactly where the sphere and the plane meet. So what you are looking at is the distance of the plane to the origin.

If the plane is given by $ax+by+cz+d$, the signed distance of a point $(x_0,y_0,z_0)$ to it is (Point-Plane Distance):

$$D = \frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}$$

which in your case gives exactly $1/\sqrt{3}$. Any point in the plane is farther to $(0,0,0)$ than $|D|$.

Answer 5

Hint:

Expand $(3x-1)^2+(3y-1)^2+(3z-1)^2\geqslant 0$ And simplify.

Answer 6

If you rewrite your proof as:

$3 ( x^2+y^2+z^2 ) \ge ( x^2+y^2+z^2 ) + ( 2xy+2yz+2zx ) = (x+y+z)^2 = 1$.

You would find that it is not so unintuitive after all.

Since you know that $x+y+z = 1$, a natural thing to do to the original equation is to homogenize it; namely make all terms have the same degree. This gives us "$3 ( x^2+y^2+z^2 ) \ge (x+y+z)^2$, and expanding out immediately tells us the solution.

In general, a technique that works for many cyclic polynomial inequalities is to try to 'smooth' the terms out. Terms that are just a power of one variable are the 'biggest', and using inequalities like "$x^2+y^2 \ge 2xy$" will 'mix' powers and thereby 'reduce' them.

Answer 7

It is enough to prove the result for $x,y,z$ positive since $$\frac{|x|+|y|+|z|}{3} \geq \frac{x+y+z}{3}$$

One can use the generalized AM-GM inequality: If $$M_p = \left(\frac{x^p+y^p+z^p}{3}\right)^{\frac{1}{p}}$$ then for $p < q$, $M_p \leq M_q$ with equality holding if and only if $x=y=z$. Here, using $M_2 \geq M_1$, we get $$\left(\frac{x^2+y^2+z^2}{3}\right)^{\frac{1}{2}} \geq \frac{|x|+|y|+|z|}{3} \geq \frac{x+y+z}{3} = \frac{1}{3}$$

Answer 8

We can minimize $x^2 + y^2 + z^2$ subject to the constraint $x+y+z = 1$ using Lagrange multipliers, we then find that $x = y = z = \frac{1}{3}$, therefore $x^2 + y^2 + z^2\geq\frac{1}{3}$.

Answer 9

One more way to prove it is by substituting out for $z$:

$$ x^2+y^2+z^2 = x^2+y^2+(1-x-y)^2=2x^2+2y^2+2xy-2x-2y+1 $$ Now, substitute $x=\hat x+a$ and $y=\hat y+b$. We get $$ 2\hat x^2+2\hat y^2+2\hat x\hat y+(4a+2b-2)\hat x+(2a+4b-2)\hat y+(2(a^2+ab+b^2-a-b)+1) $$ We can eliminate the order 1 terms by letting $4a+2b-2=2a+4b-2=0$, which gives $a=b=\frac13$. With this substitution, we have $$ 2(\hat x^2+\hat x\hat y+\hat y^2)+\frac13 $$ We can express the bracketed term as a sum of squares, giving us $$ 3(\hat x+\hat y)^2+(\hat x-\hat y)^2 + \frac13 $$ and we can see that the smallest value this can take is $\frac13$. Indeed, it takes this value when $\hat x=\hat y=0$ - that is, when $x=y=\frac13$.

Answer 10

Ok, you want an intuitive proof, not a better proof. That's fine.

First note that for $x=y=z$, we have $x^2+y^2+z^2 = 3\left(\frac13\right)^2= \frac13$. Next let us show that $x^2+y+2+z^2$ is not minimal if $x\ne y$ or $y\ne z$. Indeed if $a\ne b$ then $$ a^2 + b^2 - \left(\left(\frac{a+b}2\right)^2+\left(\frac{a+b}2\right)^2\right) = \frac12\left( a^2 + b^2 -2ab \right)=\frac12(a-b)^2>0, $$ so we can replace two of the numbers be their averages and lower the sum of squares.

Unfortunately we not yet quite done. We are done if we can show that there is a global minimum of $x^2+y^2+z^2$, given that $x+y+z=1$. Because then that global minimum must also be a local minimum, but for that we cannot have $x\ne y$ or $y\ne z$, so $x=y=z$ must be the global minimum, and its value is $\frac13$.

For that we can first argue that we can restrict ourselves to $x,y,z\ge0$ (e.g. if $x<0$ replace $(x,y,z)$ by $\frac1{-x+y+z}(-x,y,z)$, which yields a smaller value since $-x+y+z = 1 - 2x > 1$) and then appeal to the compactness of $\{(x,y,z)\colon \text{$x+y+z=1$, $x,y,z\ge0$}\}$.

I did not say that this would be pretty, but it is pretty intuitive to me :)

Answer 11

Here's another way to approach this. It's easy to see that the value of $\frac13$ is obtained when each of $x, y, z$ is $\frac13$. We want to show that as the variables deviate from this point (with their sum still being 1) the value cannot decrease.

So we look at the deviations from $\frac13$: $x=\frac13+\epsilon_1$, $y=\frac13+\epsilon_2$, $z=\frac13+\epsilon_3$ with $\epsilon_1+\epsilon_2+\epsilon_3=0$. you have

$x^2+y^2+z^2=\\ (\frac13+\epsilon_1)^2+(\frac13+\epsilon_2)^2+(\frac13+\epsilon_3)^2=\\\left(\frac19+\frac23\epsilon_1+\epsilon_1^2\right)+\left(\frac19+\frac23\epsilon_2+\epsilon_2^2\right)+\left(\frac19+\frac23\epsilon_3+\epsilon_3^2\right)=\\ \left(\frac19+\frac19+\frac19\right)+\frac23(\epsilon_1+\epsilon_2+\epsilon_3)+(\epsilon_1^2+\epsilon_2^2+\epsilon_3^2)=\\ \frac13+(\epsilon_1^2+\epsilon_2^2+\epsilon_3^2) \ge \frac13$

Answer 12

The desired inequality follows by noting that from the convexity of $f(t) = t^{2},$ we have $ f\left(\dfrac{x+y+z}{3}\right) \leq \dfrac{f(x)+f(y)+f(z)}{3}$