<s>[INST] Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$
My try:
we have $$\frac{13}{32}=\frac{2^2+3^2}{2^5}=\frac{1}{8}\left(1+(1.5)^2)\right)$$
Let $x=1.5$
Now consider the function $$f(x)=\frac{1+x^2}{8}-\ln x$$
$$f'(x)=\frac{x}{4}-\frac{1}{x}$$ So $f$ is Decreasing in $(0,2)$
any help here?
 [/INST] We'll prove that $\ln\frac{3}{2}<\frac{13}{32},$ for which we'll prove that for any $x\geq1$ the following inequality holds.
$$\ln{x}\leq(x-1)\sqrt[3]{\frac{2}{x^2+x}}.$$
Indeed, let $f(x)=(x-1)\sqrt[3]{\frac{2}{x^2+x}}-\ln{x}.$ 
Thus,
$$f'(x)=\frac{\sqrt[3]2(x^2+4x+1)-3\sqrt[3]{x(x+1)^4}}{3\sqrt[[3]{(x^2+x)^4}}=\frac{2(x^2+4x+1)^3-27x(x+1)^4}{someting\\positive}=$$
$$=\frac{(2x^2+5x+2)(x-1)^4}{someting\\positive}\geq0,$$
which gives $$f(x)\geq f(1)=0.$$
Thus, $$\ln1.5<0.5\sqrt[3]{\frac{2}{3.75}}=\frac{1}{\sqrt[3]{15}}.$$
Id est, it's enough to prove that:
$$\frac{1}{\sqrt[3]{15}}<\frac{13}{32}$$ or
$$32768<32955$$ and we are done!
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