# Comparing some numbers $a=2016^\sqrt{2014}, b=2015^\sqrt{2015},c=2014^\sqrt{2016}$

compare $a=2016^\sqrt{2014}, b=2015^\sqrt{2015},c=2014^\sqrt{2016}$

I took 2 functions to solve this question:

$f(x)=\sqrt{x+1}\ln(x-1)-\sqrt{x}\ln x$ and $g(x)=\sqrt{x}\ln x - \sqrt{x-1}\ln(x+1).$

$f(x)$ has a local maxima at a point of abscissa $x_1\in(62,63)$ and $\lim_{x\to\infty}f(x)=0$ and $f'(x)<0$ on $[x_1,\infty).\implies f(2015)>0\implies c>b.$

$g(x)$ has a local maxima in the point of absicssa $x_2\in(45,46), \lim_{x\to\infty}g(x)=0$ and $g'(x)<0$ on $[x_2,\infty)\implies g(2015)>0\implies b > a.$

Leaving us with the answer $c > b > a.$

However, I was thinking of a way of verifying this (comparing these numbers) using only $1$ function and I was thinking maybe something like $f(x)=\sqrt{x}\ln(4030+x)...$ but it doesn't really work out... how could i compare these numbers using only one function?

• is the function $f(x)=\sqrt{x}\ln(4030-x)$ instead? – Ginger88895 Jun 7 '18 at 7:08
• $f(x)= x^{\frac1{\sqrt x}}$ – Lord KK Jun 7 '18 at 7:19

Consider $f(x)=\sqrt{x}\ln(4030-x)$. It is an elementary function, so it is differentiable in $[{2000,3000}]$. Now, $f'(x)=\frac{1}{2\sqrt{x}}\ln(4030-x)-\frac{\sqrt{x}}{4030-x}$.

Intuitively, $\ln(4030-x)$ is way larger than $2$ around $x=2015$, at the same time $\sqrt{x}/(4030-x)\approx \sqrt{x}$ there, so $f'(x)>0$ around $x=2015$, therefore, $c=e^{f(2016)}>b=e^{f(2015)}>a=e^{f(2014)}$, by the monotonicity of the exponential function.

The second paragraph can be made rigorous if required...

• Can you explain a little bit more the second paragraph? and what is the intuition behind $4030$? – C. Cristi Jun 7 '18 at 7:51
• I think you mean "rigorous" (not vigorous) although I don't judge... :) – stuart stevenson Jun 7 '18 at 12:57
• @C.Cristi: $2014+2016=2015+2015=2016+2014=4030$, so the three numbers fit well – Ginger88895 Jun 8 '18 at 9:01
• @stuart stevenson: oops! I will correct it... :P – Ginger88895 Jun 8 '18 at 9:05

Take function, $f(x)= x^{\frac1{\sqrt x}}$

$f'(x) = x^{\frac1{\sqrt x}}\times {\frac{2+lnx}{2x^{5/2}}} > 0$

So, $f(2016) > f(2014) \implies 2016^{\sqrt{2014}} > 2014^{\sqrt{2016}}$

Now, take $g(x) = x^{\frac{1}{\sqrt {x-1}}}$ and $h(x) = x^{\frac{1}{\sqrt {x+1}}}$ and do the same as above.

Let $f(x)=\frac{\ln(x+1)}{\sqrt{x}}$, where $x>2000$.

Thus, $$f'(x)=\frac{\frac{\sqrt{x}}{x+1}-\frac{\ln(x+1)}{2\sqrt{x}}}{x}=\frac{\frac{2x}{x+1}-\ln(x+1)}{2x\sqrt{x}}<\frac{2-\ln(x+1)}{2\sqrt{x^3}}<0,$$ which says $$f(2015)<f(2014)$$ or $$\frac{\ln2016}{\sqrt{2015}}<\frac{\ln2015}{\sqrt{2014}}$$ or $$2016^{\sqrt{2014}}<2015^{\sqrt{2015}}.$$ Now, let $g(x)=\frac{\ln{x}}{\sqrt{x+1}}$, where $x>2000$.

Thus, $$g'(x)=\frac{\frac{\sqrt{x+1}}{x}-\frac{\ln{x}}{2\sqrt{x+1}}}{x+1}=\frac{\frac{2x+2}{x}-\ln{x}}{2(x+1)\sqrt{x+1}}<\frac{3-\ln{x}}{2\sqrt{(x+1)^3}}<0,$$ which says $$g(2015)<g(2014)$$ or $$\frac{\ln2015}{\sqrt{2016}}<\frac{\ln2014}{\sqrt{2015}}$$ or $$2015^{\sqrt{2015}}<2014^{\sqrt{2016}},$$ which says $$a<b<c.$$