Problem:
If $\lambda>1$, then $\forall p>0$, $\exists x_0 \geq0$ such that $$\lambda^{x^p} \geq x$$ $\forall x \geq x_0$.
Attempt:
Consider function $l(x)=x^p \ln \lambda - \ln x$. We can see that $l$ has a global minimum at $x^* = (\ln \lambda ^p)^{-\frac{1}{p}}$. If $l(x^*) \geq 0$, then $l(x) \geq 0$ $\forall x \geq 0$ since $x^*$ is the global minimum. If $l(x^*) < 0$, then I try to show that eventually, there is a point $x_0 > x^*$ such that $l(x_0)=0$.
At first, I was thinking that $l$ is an increasing function, but increasing does not mean that the function will surpass $0$. It is possible that the function increases and approaches $0$ but never surpasses it. However, I think that this is not possible because the derivative $l'(x)=x^{p-1}\ln \lambda ^p-\frac{1}{x}$ shows that when $x$ increases, the second term decreases while the first term increases, hence $l'$ increases overall. This means that the gradient of the curve increases. But I do not know how to write this analytically.
Kindly help me in this problem. Thank you.
$\lambda ^{x^{p}} =e^{x^{p} \ln \lambda} \geq \frac {(x^{p})^{n}(\ln \lambda )^{n}} {n!}$ for any positive integer $n$. Choose $n$ so large that $np>1$. Then $\lambda ^{x^{p}} \geq x (x^{np-1} (\ln \lambda)^{n}) \frac 1 {n!}$. Clearly $(x^{np-1} (\ln \lambda)^{n}) \geq {n!}$ if $x$ is large enough. To be precise you can take $x_0=(\ln \lambda)^{-\frac n {np-1}} {(n!)}^{\frac 1 {np-1}}$.