Finding the minimum value of $a^2+b^2+c^2$ 
Let $a$, $b$ and $c$ be $3$ real numbers satisfying $2 \leq ab+bc+ca$. Find the minimum value of $a^2+b^2+c^2$. 

I've been trying to solve this, but I don't really know how to approach this. I thought of $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca)$, but that gives me  $a+b+c$, which is unknown. How can I solve this?
 A: Famke's answer is the simplest; however, we can use a variational argument, as well.
For all variations that maintain $ab+bc+ca=2$, we have
$$
(b+c)\,\delta a+(a+c)\,\delta b+(a+b)\,\delta c=0\tag{1}
$$
To minimize $a^2+b^2+c^2$, we must have
$$
2a\,\delta a+2b\,\delta b+2c\,\delta c=0\tag{2}
$$
for all variations that satisfy $(1)$.
Orthogonality says that to have $(2)$ for all variations that satisfy $(1)$, we need
$$
2a=\lambda(b+c),\quad2b=\lambda(a+c),\quad\text{and}\quad2c=\lambda(a+b)\tag{3}
$$
Summing these up, we get that $\lambda=1$, and then solving the equations, we get that $a=b=c$. Finally, to satisfy the constraint for $(2)$, we get that $a=b=c=\sqrt{\frac23}$. Thus, the minimum of $a^2+b^2+c^2$ is $2$.
A: Scalar product:
Let $\vec A : = (a,b,c)$, $\vec B: = (b,c,a)$.
$|\vec A \cdot \vec B| \le |A| |B| $.
$\Rightarrow$  :
$2 \le ab +bc + ca \le a^2 +b^2 + c^2$.
Equality: 
$a=b=c = \sqrt{\frac{2}{3}}$.
A: You can solve this problem by your starting step.
Indeed, by C-S
$$3(a^2+b^2+c^2)\geq(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc).$$
Thus,
$$3(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+ac+bc)$$ or
$$a^2+b^2+c^2\geq ab+ac+bc$$ and since $ab+ac+bc\geq2,$ we obtain:
$$a^2+b^2+c^2\geq2.$$
The equality occurs for $(1,1,1)||(a,b,c)$ and $ab+ac+bc=2$, 
which says that $2$ is a minimal value.
A: Notice that by Inequality of arithmetic and geometric means we know that:
$$ 
2ab \leq a^2+b^2; 
\\ 
2ac \leq a^2+c^2; 
\\
2bc \leq b^2+c^2; 
$$
so we can conclude that: 
$$ 
2\left(ab+ac+bc\right) 
\leq 
2\left(a^2+b^2+c^2\right)
\Longrightarrow 
\\ 
\ \ 
\left(ab+ac+bc\right) 
\leq 
\ \ 
\left(a^2+b^2+c^2\right)
\Longrightarrow 
\\ 
\ \ \ \ \ 
\ \ \ \ \ 
\ \ \ \ \ 
\ \ \ \  
2
\leq 
\ \ 
\left(a^2+b^2+c^2\right) 
.
$$ 

Note that this in-equlality is sharp for $a=b=c=\sqrt{\dfrac{2}{3}}$;
for which one can see the value of $a^2+b^2+c^2$ is equal to $2$. 
A: Again from Cauchy-Schwarz from a different perspective 
$\sqrt{a^2c^2}+\sqrt{b^2c^2}+\sqrt{a^2c^2} \leq \sqrt{a^2+b^2+c^2}\sqrt{a^2+b^2+c^2}$
$a^2+b^2+c^2 \ge ab+bc+ac \ge 2$
$a^2+b^2+c^2 \ge 2$ 
İnequality holds for $a=b=c=\sqrt{\dfrac{2}{3}}$
