Extrema shift

APOLOGIES. THIS PROOF CONTAINS AN ERROR AND I AM WORKING ON AN UPDATE. Let us write

$y=e^{x}-ax$

There is a real value of $y$ for every real $x$ , which means that between the two roots, there is a stationary point (zero gradient).

$dy/dx=e^x+a=0$ at the stationary point

$e^x=a$

$x=ln a$

and $a$ is therefore positive, and the curve is U-shaped tending to +inf .

$y<0$ therefore, when $x=ln(a)$ , or there would be no roots.

$e^{ln(a)}-aln(a)<0$

$e-a<0$

$a>e$

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let our roots be $x=p$ and $x=q$ . such that $p>q$ . easier to read

$e^{p}=ap$ and $e^{q}=aq$

$e^{p+q}=a^{2}pq$ ...........1 THIS IS CORRECT

$p+q=2 ln(a)+ln(p)+ln(q) > 2$ since we know $ln(a)>1$ since $a>e$ THIS IS THE ERROR

$\boxed{p+q>2}$ THIS CONCLUSION IS NOT YET JUSTIFIED

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rename the two roots $p=ln(a)+m$ and $q=ln(a)-n$ . We know that m and n are positive as the roots are either side of $x=ln(a)$

$e^{ln(a)+m}-aln(a)-am=0$ and $e^{ln(a)-n}-aln(a)+an=0$

take the difference on both sides

$e^{-n}+e^m=(n-m)$

therefore $n-m>0$

$ln(a)+m+ln(a)-n<2ln(a)$

$p+q<2ln(a)$

$e^{p+q}<a^2$ ...........2

take an equation (1) from above

$e^{p+q}=a^{2}pq$

$pq=e^{p+q}/a^{2}$

from equation 2 above we know RHS<1, therefore

$\boxed{pq<1}$

and we are done.