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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

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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.

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