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.
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Reinstate Monica
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Is $\angle DAO$ a right angle? – Michael Biro Oct 22 '15 at 01:43
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@MichaelBiro yes – Reinstate Monica Oct 22 '15 at 01:49
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$\tan(a)=\tan(b)+\tan(c)~=>~\sec(a)-1<2~\Big[\sec(b)+\sec(c)-2\Big].$ – Lucian Oct 22 '15 at 07:28
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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 $$ \sqrt{(a+b)^2+1}-1\lt2\left(\sqrt{a^2+1}-1+\sqrt{b^2+1}-1\right) $$ $\square$
robjohn
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hint:
$\dfrac{\sqrt{x}+\sqrt{y}}{2} \le \sqrt{\dfrac{x+y}{2}}$
$AD,AB,AC$ can be writen as $\sqrt{(1+d)^2-1}$
chenbai
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