Proof $\frac{ln(1+xy)}{x^2ln64}-\frac{y}{x(1+xy)ln64}>-\frac{1}{4x^2ln4}\ \forall x\in[\frac{1}{4};4]$ when $y\ge1$

Question

Proof $$\frac{\ln(1+xy)}{x^2\ln64}-\frac{y}{x(1+xy)\ln64}>-\frac{1}{4x^2\ln4}$$ for all $x\in[\frac{1}{4};4]$ when $y\ge1$

I tried to do this: $$\frac{\ln(1+xy)}{x^2\ln64}-\frac{y}{x(1+xy) \ln64}=\frac{(1+xy)\ln(1+xy)-xy}{x^2(1+xy)\ln64}=\frac{\ln(1+xy)-\frac{xy}{1+xy}}{x^2\ln64}$$ But it does not seem helpful

Answer 1

Let $xy=t$.

Thus, $t>0$ and $$\left(\ln(1+t)-\frac{t}{1+t}\right)'=\left(\ln(1+t)-\frac{t+1-1}{1+t}\right)'=\frac{1}{1+t}-\frac{1}{(1+t)^2}=\frac{t}{(1+t)^2}>0,$$ which says that $$\ln(1+xy)-\frac{xy}{1+xy}>\ln(1+0)+\frac{0}{1+0}=0.$$ Can you end it now?