First of all, sorry, that I didn't manage to write the equation with a proper editor: $$ \frac{a^x}{x} + a^\frac{1}{x}x = 2a $$ I know that $x=1$. It's easy. But I couldn't solve it.
I try two methods, but I am stuck with other equations:
First approach: $$ a^x=y \qquad \text{and} \qquad a^{1/x}=z $$ $$ x \ln{a}=\ln{y} \qquad \text{and} \qquad \ln{a}=x \ln{z} $$ so $$ x = \frac{\ln{y}}{\ln{a}}=\frac{\ln{a}}{\ln{z}} $$ so $$(\ln(y))(\ln(z))=(\ln(a))^2$$
Putting in the original equation I obtain the following system:
$$(\ln(y))(\ln(z))=(\ln(a))^2$$
and
$$z(\ln(y))+ y(\ln(z))= 2a(\ln(a))$$ or the equivalent
$$z/(\ln(z))+y/(\ln(y))= 2a/(\ln(a))$$
And if I try to use W lambert, I obtain something very... ugly.
Second approach:
$$(a^x)/x + (a^{1/x})x = 2a$$
$$(a^{1/x})(x^2)+ (a^x) - 2ax =0$$
$$\ln((a^{1/x})(x^2))= \ln(2ax - a^x)$$
$$\ln((a^{1/x}))+ 2 (\ln (x)) = \ln(2ax - a^x)$$
$$(1/x)(\ln(a)) + 2 (\ln(x))= \ln (2ax - a^x)$$
$$\ln(a) + 2 (\ln (x)) = x \ln (2ax - a^x)$$
$$\ln(a(x^2))= \ln(2ax - a^x)^x)$$
$$a(x^2)= (2ax- a^x)^x$$
If I try to use W lambert, I obtain again, an ugly equation...
Someone to help me?
Thank you.
By AM-GM,
$$\frac{a^x}{x}+\frac{a^{1/x}}{1/x} \geq 2a^{\frac{x+1/x}{2}} \geq 2a$$ provided $a > 1$, with equality iff $x+x^{-1}=2$ iff $x=1$.
If $a < 1$, by weighed AM-GM, let $s=x+x^{-1}$, then
$$LHS \geq sa^{xx^{-1}/s}a^{x^{-1}x/s} = sa^{2/s}.$$
Since $s \geq 2$ and $a < 1$, $sa^{2/s} \geq 2a$ with equality iff $s=2$ iff $x=1$.