We know if we have $n$ numbers say $a_1,a_2,.....,a_n$ then AM $\geq$ GM implies

$\frac{a_1+a_2+....+a_n}{n} \geq (a_1.a_2........a_n)^\frac{1}{n}$

I assure you that the sum the can be done just by using this simple inequality,why don't you just give it a try?

Applying AM $\geq$ GM on $a^5,a^5,a^5,b^5,c^5$ we get

$3a^5+b^5+c^5 \geq 5a^3bc$

Similarly, $a^5+3b^5+c^5 \geq 5ab^3c$

$a^5+b^5+3c^5 \geq 5abc^3$

Adding the above three equations we get $a^5+b^5+c^5 \geq abc(a^2+b^2+c^2)$

So you have all the pieces of the jigsaw puzzle with you,and the puzzle is about to be completed,just try to place the remaining few pieces in its correct position

Now we have to show that $a^2+b^2+c^2 > ab+bc+ca$

Applying AM $\geq$ GM on $a^2,b^2$ we get,

$a^2+b^2 \geq 2ab$

Similarly,$b^2+c^2 \geq 2bc$

$a^2+c^2 \geq 2ca$

Adding the above three equations we get $a^2+b^2+c^2 \geq ab+bc+ca$

We are almost there ,so just try the last step yourself.

Therefore, $a^5+b^5+c^5 \geq abc(a^2+b^2+c^2) \geq abc(a^2+b^2+c^2)$

i.e, $a^{5}+b^{5}+c^{5}>a b c(a b+b c+c a)$