I am trying to find out how to algebraically prove that $\left(\frac 53\right)^{1/2} \left(n\right)^{1/4} \geq \log n$.
The statement above is true for all $n \geq 1$, but I must find a way to prove this statement algebraically rather than just plugging values in for "$n$".
I am aware that to prove this inequality algebraically, I need to use properties of logarithms, but I am not sure which one to use. Please help. Thank you.
Your inequality doesn't hold for all $n\ge 1.$ For example try $x=54.$ Let me prove a corrected (best possible) version of it. Lets consider the function $$f(x)=\dfrac{\ln x}{x^{1/4}}, \quad x\ge1.$$ Using the derivative of this function, you can easily find that its absolute maximum is $4/e$ occurs at $x=e^4.$ Thus $$\dfrac{4}{e}x^{1/4}\ge\ln x$$ for all $x\ge 1.$
If you do not consider the previous deduction as an algebraic approach, note that for each $n\ge 1$ there is a unique $x\ge 1$ such that $n=x^4,$ which is the fourth root of $n.$ Then the inequality reduces to $\dfrac{x}{e} \ge \log x,$ and after exponentiating both sides we can get $$e^x\ge x^e.$$ This last form has been considered in this nice question.
