Note that $\frac{p^k}{q^{k-1}} \geq pk-q(k-1)$, because $p^k+q^k+q^k+...+q^k \geq kpq^{k-1}$ by AM-GM. So, $LHS \geq 5a-4x+4b-3y+3c-2z=x+y+z$ and we are done.

This problem was proposed by Burii.

By bernoulli, \[LHS=(\frac{a}{x})^5x+(\frac{b}{y})^4y+(\frac{c}{z})^3z=((\frac{a}{x}-1)+1)^5x+((\frac{b}{y}-1)+1)^4y+((\frac{c}{z}-1)+1)^3z\geq x(\frac{5a}{x}-5+1)+y(\frac{4b}{y}-4+1)+z(\frac{3c}{z}-3+1)\]\[=5a-5x+x+4b-4y+y+3c-3z+z=x+y+z \]As desired.$\blacksquare$

$LHS + 4x + 3y + 2z$ $\geq$ $5a + 4b + 3c$ hence $LHS$ $\geq$ $x + y + z$.