inequality $\left(\frac{5}{4} \right)^{\sqrt{2}}< \left(\frac{6}{5} \right)^{\sqrt{3}}$

Hint: Since $\log (x) $ is increasing function and $$x-\frac12x^2+\frac13x^3-\frac14x^4<\log (1+x)< x-\frac12x^2+\frac13x^3-\frac14x^4+\frac15x^5$$ for $x>0$, it suffices to prove: $$\frac14\left (1-\frac1{2\!\cdot\!4}+\frac1{3\!\cdot\!4^2}-\frac1{4\!\cdot\!4^3}+\frac1{5\!\cdot\!4^4}\right)\sqrt2 <\frac15\left (1-\frac1{2\!\cdot\!5}+\frac1{3\!\cdot\!5^2}-\frac1{4\!\cdot\!5^3}\right)\sqrt3. $$

The given inequality is equivalent to $$\sqrt{2}\log \frac{5}{4}<\sqrt{3}\log\frac{6}{5}$$ or $$\frac{\log(25/24)}{\log(6/5)}<\frac{\sqrt{3}-\sqrt{2}}{\sqrt{2}}$$ Now $$\log \frac{6}{5}=\log\frac{1+1/11}{1-1/11}>\frac{2}{11}$$ and hence our job is done if we can show that $$\frac{11}{2}\log\frac{25}{24}<\frac{\sqrt{6}-2}{2}\tag{1}$$ Next we note that $$\frac{1}{2}\log\frac{1+x}{1-x}=\sum_{n\geq 0}\frac{x^{2n+1}}{2n+1}<x+\frac{x^3}{3(1-x^2)}$$ Putting $x=1/49$ we see that LHS of above inequality is $(1/2)\log(25/24)$ and thus $$\frac{11}{2}\log(25/24)<\frac{11}{49}+\frac{11}{3\cdot 49}\frac{1}{49^2-1}$$ The right side can be evaluated easily using division by small primes $2,3,5,7$ to get $0.2245$. The right side of $(1)$ is also easily evaluated to $0.2247$.

Since $\dfrac54=\dfrac65+\dfrac{1}{20}$ I will solve this problem with a more general context.

Let the curves $f(x)=\left(x+\dfrac{1}{20}\right)^{\sqrt2}$ and $g(x)=x^{\sqrt3}$ so we want to prove $f(\frac65)\lt g(\frac65)$

These curves are convex and solving the equation $$\left(x+\dfrac{1}{20}\right)^{\sqrt2}=x^{\sqrt3}$$ there is only one solution $x_0\approx 1.19931$

We can verify that, for example $$f(\frac{19}{20})\gt g(\frac{19}{20})\\f(x_0)=g(x_0)\\ f(\frac{39}{20})\lt g(\frac{39}{20})$$ Therefore because the curves are convex we must have $$f(x)\gt g(x)\hspace{1cm}\text { for } x\lt x_0 \\f(x_0)=g(x_0)\hspace{1cm}\text { for } x=x_0\\ f(x)\lt g(x)\hspace{1cm}\text { for } x\gt x_0$$ Thus since $\dfrac 65=1.2\gt x_0=1.19931$ we have $f(x)\lt g(x)$

In particular we have as a nice example $(\pi+\frac{1}{20})^{\sqrt2}\lt (\pi)^{\sqrt3}$