Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$ How to prove this for positive real numbers?
$$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$
I tried AM-GM, CS inequality but all failed.
 A: I have come up with an answer with myself. Using CS inequality
$$(a^4+b^4+c^4)(1+1+1)\geq(a^2+b^2+c^2)^2$$
$$(a^2+b^2+c^2)(1+1+1)\geq(a+b+c)^2$$
Hence we have
$$a^4+b^4+c^4\geq\frac{(a+b+c)^4}{27}=(a+b+c)\left(\frac{a+b+c}{3}\right)^3\geq abc(a+b+c)$$
A: Here other two answers used Cauchy-Scwartz Inequality. I am giving a simple $AM\ge GM$ inequality proof.
You asked, $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c\\\implies a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$$
Now, from, $AM\ge GM$, we have $$\frac {a^4+
a^4+b^4+c^4}4\ge \left(a^4\cdot a^4\cdot b^4\cdot c^4\right)^{1/4}=a^2bc\tag 1$$
Similarly, $$\frac {a^4+
b^4+b^4+c^4}4\ge \left(a^4\cdot b^4\cdot b^4\cdot c^4\right)^{1/4}=ab^2c\tag 2$$ and also, $$\frac {a^4+
b^4+c^4+c^4}4\ge \left(a^4\cdot b^4\cdot c^4\cdot c^4\right)^{1/4}=abc^2\tag 3$$
Now, summing up $(1),(2),(3)$, we have, $a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$, that is $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c$$
A: By Holder 
$$\sum_{cyc}\frac{a^3}{bc}\geq\frac{(a+b+c)^3}{3(ab+ac+bc)}=\frac{(a+b+c)\cdot(a+b+c)^2}{3(ab+ac+bc)}\geq a+b+c$$
A: Using Cauchy-Schwarz Inequality twice:
$a^4 + b^4 +c^4 \geq a^2b^2 +b^2c^2 +c^2a^2 \geq ab^2c +ba^2c +ac^2b = abc(a+b+c)$
A: Another way.
By Rearrangement 
$$\sum_{cyc}\left(\frac{a^3}{bc}-a\right)=\sum_{cyc}\frac{a}{bc}(a^2-bc)\geq\frac{1}{3}\sum_{cyc}\frac{a}{bc}\sum_{cyc}(a^2-bc)=\frac{1}{6}\sum_{cyc}\frac{a}{bc}\sum_{cyc}(a-b)^2\geq0.$$
