Proof of one inequality $a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$

Question

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.

Answer 1

Using Cauchy-Schwarz Inequality twice:

$a^4 + b^4 +c^4 \geq a^2b^2 +b^2c^2 +c^2a^2 \geq ab^2c +ba^2c +ac^2b = abc(a+b+c)$

Answer 2

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 $$a^4+b^4+c^4\geq\frac{(a+b+c)^4}{27}=(a+b+c)\left(\frac{a+b+c}{3}\right)^3\geq abc(a+b+c)$$