If $a,b,c$ are positive real numbers prove that $$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$$
I state the following:
Multiplying both sides by $abc$ yields $$(abc)(a+b+c)?{a^4+b^4+c^4}$$
By AM-GM for $abc$ we have $$abc\leq\left(\frac{a+b+c}{3}\right)^3\Rightarrow\left(\frac{a+b+c}{3}\right)^3(a+b+c)?{a^4+b^4+c^4}$$
This can be rewritten as the following:
$$\left(\frac{a+b+c}{3}\right)^4\leq\frac{a^4+b^4+c^4}{3}$$
which holds due to the power mean inequality.
I was also thinking that it may be proven by a double application of the Chebyshev inequality as well
To begin, hold that ${a}\ge{b}\ge{c}$ observe that $$\left[\begin{matrix} a^3 & b^3 & c^3 \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\end{matrix}\right]=a^2+b^2+c^2\leq\left[\begin{matrix} a^3 & b^3 & c^3 \\ \frac{1}{b} & \frac{1}{c} & \frac{1}{a}\end{matrix}\right]=\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}$$
Thus we can hold confidently even if we assume equality that $$\left[\begin{matrix} {a^2} & {b^2} & c^2 \\ \frac{1}{a} & \frac{1}{b} & \frac{1}{c}\end{matrix}\right]=a+b+c\leq\left[\begin{matrix} \frac{a^3} {b} & \frac{b^3}{c} & \frac{a^3}{a} \\ \frac{1}{c} & \frac{1}{a} & \frac{1}{b}\end{matrix}\right]=\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab}$$
Please provide feedback if you can.
Your idea is correct, but the proof is a bit confusing because it is not always clear if you assume the desired inequality or not. I would write it as follows:
Using the AM-GM inequality $(1)$ and the power-mean inequality $(2)$ we have $$ abc(a+b+c)\underset{(1)}{\leq}(a+b+c)\left(\frac{a+b+c}{3}\right)^3 = 3\left(\frac{a+b+c}{3}\right)^4 \underset{(2)}{\leq} a^4 + b^4 + c^4 $$ and dividing by $abc$ gives $$ a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ac}+\frac{c^3}{ab} \, . $$