algebra precalculus - Solution of equation $frac{xcdot 2014^{frac{1}{x}}+frac{1}{x}cdot 2014^x}{2} = 2014$

Solution of equation $\displaystyle \frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014$

$\bf{My\; Try::}$ Clearly Here $x>0$, Now Using $\bf{A.M\geq G.M}$

So Here $\displaystyle x\cdot 2014^{\frac{1}{x}}>0$ and $\displaystyle \frac{1}{x}\cdot 2014^x>0$ .

So $\displaystyle \left(\frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2}\right)\geq \left(x\cdot 2014^{\frac{1}{x}}\cdot \frac{1}{x}\cdot 2014^x\right)^{\frac{1}{2}} = \sqrt{2014^{\left(x+\frac{1}{x}\right)}} = 2014$

And equality Hold , When $\displaystyle x = \frac{1}{x}\Rightarrow x= 1$

Can we solve it without Using $\bf{A.M\geq G.M}$

If Yes, Then please explain me.

Thanks

Answer

Applying AM-GM probably gives the most elegant solution (and indeed the presentation is quite suggestive of AM-GM), but this can also be solved by using an idea behind AM-GM. Noting first that $x\ge0$, we have:

$$\begin{align}
&\frac{x\cdot 2014^{\frac{1}{x}}+\frac{1}{x}\cdot 2014^x}{2} = 2014
\\\implies&x2014^{\frac{1}{x}-1}+\frac{1}{x}2014^{x-1}=2
\\\implies&\left(\sqrt{x2014^{\frac{1}{x}-1}}-\sqrt{\frac{1}{x}2014^{x-1}}\right)^2+2\sqrt{2014^{\frac{1}{x}+x-2}}=2
\\\implies&\left(\sqrt{x2014^{\frac{1}{x}-1}}-\sqrt{\frac{1}{x}2014^{x-1}}\right)^2+2\sqrt{2014^{\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2}}=2

\\\implies&2\sqrt{2014^{\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2}}\le2
\\\implies&2014^{\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2}\le1
\\\implies&\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2\le0
\\\implies&\left(\sqrt{\frac{1}{x}}-\sqrt{x}\right)^2=0
\\\implies&\sqrt{\frac{1}{x}}-\sqrt{x}=0
\\\implies&\frac{1}{x}=x
\\\implies&x=1
\end{align}$$

where we have used the fact that $x\ge 0$ several times.