Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$ Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.
Remember, that for any nonzero $x$, one has

$$
x+\frac{1}{x} \geq 2 \tag{1}
$$

We apply this inequality repeatedly. Put

$$
T=\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab}
$$

We have 
$$
T=\frac{a^2+b^2}{abc}+\frac{c}{ab} =\frac{\sqrt{a^2+b^2}}{ab} \bigg(\frac{\sqrt{a^2+b^2}}{c}+\frac{c}{\sqrt{a^2+b^2}
}\bigg) \geq \frac{2(\sqrt{a^2+b^2})}{ab},
$$

by using (1) with $x=\frac{\sqrt{a^2+b^2}}{c}$. Then,

$$
T \geq \frac{2}{\sqrt{ab}} \sqrt{\frac{a}{b}+\frac{b}{a}} \geq \frac{2\sqrt{2}}{\sqrt{ab}}
$$
by using (1) with $x=\frac{a}{b}$. So $T \geq \sqrt{\frac{8}{ab}}$. 

We see that if $ab \leq 8$, we are done. So we may assume $ab \geq 8$. By symmetry, we may also assume
$ac \geq 8, bc \geq 8$. But then

$$
1 \leq \dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \leq \frac{1}{8}+\frac{1}{8}+\frac{1}{8} =\frac{3}{8},
$$

which is impossible.
multiply abc

$a+b+c\geq abc $

$a^2+b^2+c^2\geq abc $

$a^2+b^2+c^2-(a+b+c)$

$⇔(a-\frac12)^2+(b-\frac12)^2+(c-\frac12)^2\geq \frac34$

therefore

$a^2+b^2+c^2\geq \frac34+abc$