This other question includes the following strengthened version of the arithmetic-mean geometric-mean inequality.
\begin{equation} \label{1}\tag{1} \dfrac{a+b}{2} - \sqrt{ab} \geq \dfrac{1}{16 \max \left\lbrace a , b \right\rbrace} \left( a - b \right) ^{2} . \end{equation}
I'm wondering if anyone has a 1) proof of this and 2) a citation for it? My initial thought was to use Lagrange multipliers here, but the max on the bottom makes this approach a bit annoying.
Here is an image version of the solution
Written out here:
We first break symmetry by assuming that $a \geq b$. We need to show then that $\frac{a+b-2\sqrt{ab}}{2} \geq \frac{\left(a-b\right)^2}{16a}.$
This is equivalent after clearing denominates and factoring the right hand side to $$8a(\sqrt{a}-\sqrt{b})^2 \geq (a-b)^2$$ which is equivalent to $$8a\frac{(\sqrt{a}-\sqrt{b})^2}{(a-b)^2} \geq 1$$ which is equivalent to $$8a\frac{1}{(\sqrt{a}+\sqrt{b})^2} \geq 1.$$
So we are done if we can show that $8a \geq (\sqrt{a}+\sqrt{b})^2.$
By the usual AM-GM inequality in the form $(x+y)^2 \leq2(x^2 +y^2)$, and using that $a \geq b$ we have $8a \geq 4a \geq 2(a+b) \geq (\sqrt{a}+\sqrt{b})^2$ and we are done. Note that this proof also shows that we could get a stronger constant in the original, replacing $16$ with $8$ in our original inequality.