Inequality. $a^2+b^2+c^2 \geq a+b+c$

Using Cauchy-Schwarz inequality we get $$a+b+c=a\cdot 1+b\cdot 1+c\cdot 1+\leq\sqrt{a^2+b^2+c^2}\sqrt{1^2+1^2+1^2}\tag{1}$$ From AM-GM we obtain $a^2+b^2+c^2\geq 3\sqrt[3]{a^2b^2c^2}=3$, so $$\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\tag{2}$$ From $(1)$ and $(2)$ it follows $$a+b+c\leq\sqrt{a^2+b^2+c^2}\sqrt{3}\leq\sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}=a^2+b^2+c^2$$

Let's solve it in an elementary way and start from the fact that: $$a^2 \ge 2a -1 \tag1$$ $$b^2 \ge 2b-1 \tag2$$ $$c^2 \ge 2c-1 \tag3$$

Then add up $(1)$ $(2)$ $(3)$ and get: $$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \tag4$$ By AM-GM we have $$\frac{a+b+c}{3} \ge (abc)^\frac{1}{3}=1$$ $$a+b+c \ge 3 \tag5$$ Finally, from $(4)$ and $(5)$ we obtain the required inequality: $$a^2+b^2+c^2 \ge a+b+c +a+b+c -3 \ge a+b+c$$

Q.E.D.

As A.M. of any set of positive number ≥ G.M,

$\frac{a+b+c}{3}≥ (abc)^{\frac{1}{3}}$

Now, $3(a^2+b^2+c^2)-(a+b+c)^2= (a-b)^2+(b-c)^2+(c-a)^2≥ 0$

=>$a^2+b^2+c^2≥ \frac{(a+b+c)^2}{3}$

=>$a^2+b^2+c^2≥(a+b+c)(\frac{a+b+c}{3})≥a+b+c$ as $\frac{a+b+c}{3}≥1$