geometric inequality involving lengths of hypotenuses In the diagram, $AD = AB + AC$ and all these lengths are positive. The following inequality holds:
$$DG < 2(BE + CF).$$
The constant 2 is sharp. I am able to prove the weaker inequality $DG < BE + 3 CF$ where $CF \ge BE$ but I am not able to prove the stronger inequality. Could someone please suggest an approach? I am not looking for a complete answer.


Using $a=AB$, $b=AC$, and $a+b=AD$, this is equivalent to the following
Lemma: For $a,b\gt0$,
$$
\sqrt{(a+b)^2+1}-1\lt2\left(\sqrt{a^2+1}-1+\sqrt{b^2+1}-1\right)
$$
Proof: Since $2ab\le a^2+b^2$, we have $(a+b)^2\le2\left(a^2+b^2\right)$. Therefore,
$$
\frac{(a+b)^2}{\sqrt{(a+b)^2+1}}\lt2\left(\frac{a^2}{\sqrt{a^2+1}}+\frac{b^2}{\sqrt{b^2+1}}\right)
$$
Substituting $a\mapsto at$ and $b\mapsto bt$, multiplying by $\frac2t$, and integrating from $0$ to $1$ yields
$$
\sqrt{(a+b)^2+1}-1\lt2\left(\sqrt{a^2+1}-1+\sqrt{b^2+1}-1\right)
$$
$\square$
