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 $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.
 A: 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.
A: Denote $X=\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}$ and $Y=\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}$. In fact, if $X\ge 1$, then $Y\ge \sqrt{3}$. ($\sqrt{3}$ is optimal, because when $a=b=c=\frac{1}{\sqrt{3}}$, $X=1$ and $Y=\sqrt{3}$.)
By Cauchy-Schwarz inequality,
$$3(a^2+b^2+c^2)\ge (a+b+c)^2.$$
It follows that
$$Y=\frac{a^2+b^2+c^2}{abc}\ge \frac{(a+b+c)^2}{3abc}=\frac{abc}{3}  X^2\ge \frac{abc}{3}.$$
If $abc\ge 3\sqrt{3}$, we are done. Otherwise, by inequality of arithmetic and geometric means,
$$Y\ge 3(abc)^{-\frac{1}{3}}\ge \sqrt{3}.$$
A: I want to give a complete answer. 
Assume by the sake of contradiction that $$\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}<1.\tag{1}$$
By Cauchy Schwarz it follows from $(1)$ that $$\frac{3}{abc}>\frac{3}{abc}\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\geq \left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)^2.\tag{2}$$
From $(2)$ we recover two facts


*

*By $AM-GM$ mean $$\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)^2\geq 3\left(\frac{1}{a^2bc}+\frac{1}{ab^2c}+\frac{1}{abc^2}\right),$$ therefore, combined with $(2)$, we obtain on one hand $$1>\frac{1}{a}+\frac{1}{b}+\frac{1}{c},\tag{3}$$ which implies that $\min\{a,b,c\}>1$ (remember that they are all positive).

*On the other hand we can also derive from $(2)$ and $(3)$ the following: $$\frac{1}{\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)}>\frac{a+b+c}{3}>1,\tag{4}$$


which in the end leads to an absurd with respect to the hypothesis of the problem.
Then $(1)$ must be false and the problem is solved.
A: 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
$a^2+b^2+c^2\geq \frac34+abc$
