Let $x+y+z = 5$ where $x,y,z \in \mathbb{R}$, prove that $x^2+y^2+z^2\ge \frac{5}{3}$ My thinking:
Since $x+y+z = 5$, we can say that $x+y+z \ge  5$.
By basic fact: $x^2,y^2,z^2\ge 0$
If $x+y+z \ge  5$, then $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$
If $\frac{\left(x+y+z\right)}{3}\ge \frac{5}{3}$ and $x^2,y^2,z^2\ge 0$, hence $\frac{x^2+y^2+z^2}{3}\ge \frac{5}{3}$...???
I'm not sure about this, can someone help?
 A: Viewed geometrically, $x+y+z=5$ is a plane and $x^2+y^2+z^2=a^2$ is a sphere with radius $a$ at the origin; the smallest (positive) $a$ for which the sphere touches the plane corresponds to the shortest distance between the plane and the origin, and the point of tangency represents the smallest possible value of $x^2+y^2+z^2$. Intuitively this is achieved at $x=y=z=\frac53$, whence the result follows (indeed $x^2+y^2+z^2\ge\frac{25}3$).
A: I would preface by saying that we can prove a stronger inequality, namely $x^2+y^2+z^2 \geq \frac{25}3$.
You can use the Cauchy-Schwarz inequality. For three variables, it states that $(a_1b_1+a_2b_2+a_3b_3)^2 \leq (a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)$. Using $a_1=a_2=a_3=1$ and $b_1=x,b_2=y,b_3=z$ gives $25\leq 3(x^2+y^2+z^2)$.
You can also use Lagrange multipliers. We get $(2x,2y,2z)=\lambda(1,1,1)$, so that an extreme value of $x^2+y^2+z^2$ is achieved when $x=y=z$. If $x+y+z=5$, this means $x=y=z=\frac53$, giving $x^2+y^2+z^2 = \frac{25}3$. To get the inequality in the right direction, we notice that if $x=y=0$ and $z=5$ for example, then $x^2+y^2+z^2=25\geq \frac{25}3$.
A: $(x-1)^2+(y-1)^2+(z-1)^2\ge 0$. Hence $x^2+y^2+z^2-2(x+y+z)+3\ge 0$, so $x^2+y^2+z^2\ge 10-3=7>5/3$. Moreover $(x^2+y^2+z^2)/3\ge 7/3>5/3$
Your proof is not justified: the last "hence" statement does not follow from the previous statements.
