Prove or disprove
let $n>4$ be give postive integers,if $a$ is also be give postive integers,such $\sqrt[n]{a}$ be not postive integers,show that: for any postive $m$,there exist constant $c$,such $$\{m\sqrt[n]{a}\}>\dfrac{c}{m^{n-2}}$$ where $\{x\}=x-[x]$
My try: let $[m\sqrt[n]{a}]=k$,then $m\sqrt[n]{a}>k$,so we have $$m^n\cdot a-k^n\ge 1$$ so we have $$\{m\sqrt[n]{a}\}=(m\sqrt[n]{a}-k)\ge\dfrac{1}{(m\sqrt[n]{a})^{n-1}+(m\sqrt[n]{a})^{n-2}k+\cdots+k^{n-1}}>\dfrac{1}{n(m(\sqrt[n]{a})^{n-1}}=\dfrac{c}{m^{n-1}}$$ where $c=\dfrac{1}{n(\sqrt[n]{a})^{n-1}}$ But I can't prove the $LHS>\dfrac{c}{m^{n-2}}$