Proving that $x^{2x}<(2x-1)^{2x-1}$

Question

I came across a problem that asks me to prove that $\log_2(3) > \log_3(5)$ and I ended up needing to prove that : $$ x^{(2x)} < (2x-1)^{(2x-1)}$$ for $x > 1$. I tried to solve it but i couldn't. Do you have any suggestions ?

Answer 1

I think, you need to prove that: $$\ln^2(2n-1)>\ln{n}\ln(2n+1)$$ for any natural $n\geq2$.

For any $n\geq3$ it's true by AM-GM: $$\ln{n}\ln(2n+1)\leq\left(\frac{\ln{n}+\ln(2n+1)}{2}\right)^2=\ln^2\sqrt{2n^2+n}<\ln^2(2n-1).$$ By the way, your inequality follows from the following. $$\log_23>\frac{3}{2}>\log_35.$$ The left inequality it's $$3>2^{\frac{3}{2}}$$ or $$9>8.$$ The right inequality it's $$3^{\frac{3}{2}}>5$$ or $$27>25.$$

Answer 2

Then $x^{2x} < (2x-1)^{(2x - 1)}$ would follow from $x^{2x} < (2x)^{(2x - 1)}$, so attempting

$$x^{2x} < (2x)^{(2x - 1)}$$ $$\log_2 \left(x^{2x}\right) < \log_2\left((2x)^{(2x - 1)}\right)$$ $$2x\log_2 x < (2x - 1) \left(\log_2 2 + \log_2 x\right)$$

Can you finish?

Answer 3

Much simpler approach to the original problem:

$3^2>2^3 \Rightarrow \log_2(3) > \frac 3 2 \\5^2<3^3 \Rightarrow \log_3(5) < \frac 3 2 \\ \Rightarrow \log_2(3) > \log_3(5)$