Bulgarian National Mathematical Olympiad

We will prove that if $x>-1$, $x\ne 0$, and $\alpha \in (1,2)$, then $$(1)0<f(x)=(1+x{)}^{\alpha }-1-\alpha x-\frac{\alpha (\alpha -1)}{2}{x}^2-\frac{\alpha (\alpha -1)(\alpha -2)}{6}{x}^3.$$ We have $$f^{\prime }(x)=\alpha \left[(1+x{)}^{\alpha -1}-1-(\alpha -1)x-\frac{(\alpha -1)(\alpha -2)}{2}{x}^2\right],$$ $$f^{\prime \prime }(x)=\alpha (\alpha -1)\left[(1+x{)}^{\alpha -2}-1-(\alpha -2)x\right],$$ $$f^{\prime \prime \prime }(x)=\alpha (\alpha -1)(\alpha -2)\left[(1+x{)}^{\alpha -3}-1\right],$$ Since $f^{\prime \prime \prime }(x)<0$ for $x\in (-1,0)$ and $f^{\prime \prime \prime }(x)>0$ for $x>0$, we have $f^{\prime \prime }(x)>f^{\prime }(0)=0$ for $x>-1$, $x\ne 0$. Then $f^{\prime }(x)<f^{\prime }(0)=0$ for $x\in (-1,0)$ and $f^{\prime }(x)>f^{\prime }(0)>0$ for $x>0$, whence $f(x)>f(0)=0$ for $x>-1$, $x\ne 0$.

We now prove that ${\left(\frac{6}{5}\right)}^{\sqrt{3}}>{\left(\frac{5}{4}\right)}^{\sqrt{2}}$. Set $x=\frac{1}{5}$ and $\alpha =\sqrt{\frac{3}{2}}$. According to (1), it is enough to check that $$\alpha x+\frac{\alpha (\alpha -1)}{2}{x}^2+\frac{\alpha (\alpha -1)(\alpha -2)}{6}{x}^3>\frac{1}{4}\iff$$ $$\frac{277}{1500}\alpha +\frac{3}{125}>\frac{1}{4}\iff \alpha >\frac{339}{277}\iff 3.277^2>2.339^2\iff 230187>229842.$$ The last inequality is obvious and this completes the solution.