A proposition says that the sum of the sides of a quadrilateral is greater than the sum of its diagonals. The proof for this proposition goes as follows:-
In quadrilateral $\mathrm{ABCD,}$ we have in $\Delta$$\mathrm{ACD,}$ $\mathrm{AC<AD+CD⋯(1)}$
Similarly,
In $\Delta$$\mathrm{ACB,}$
$\mathrm{AC<AB+CB⋯(2)}$
In $\Delta$$\mathrm{BCD,}$
$\mathrm{BD<BC+CD⋯(3)}$
and, in $\Delta$$\mathrm{ADB,}$
$\mathrm{BD<AD+AB⋯(4)}$
Hence, on adding all the equations, we achieve
$\mathrm{2(AB+BC+CD+AD)>2(AC+BD)}$
i.e., $\mathrm{AB+BC+CD+DA>AC+BD.}$
This proves our proposition well and good but out of nowhere I got stuck on an another seemingly strange but obvious thought process that should be an another proof to the proposition but somehow I guess I missed something important in the proof that contradicts the final results. Described below is the same deficient proof I am stuck upon.
In the same quadrilateral $\mathrm{ABCD,}$ diagonals $\mathrm{AC}$ and $\mathrm{BD}$ meet each other at $\mathrm{O.}$ Hence, they form the triangles $\mathrm{AOD , AOB , BOC}$ and $\mathrm{DOC.}$
Thence, on applying the triangle inequalities in the above triangles, we get
In $\Delta$$\mathrm{AOD}$, we have $\mathrm{AO + OD > AD...(i)}$
In $\Delta$$\mathrm{AOB}$, we have $\mathrm{AO + BO > AB...(ii)}$
In $\Delta$$\mathrm{BOC}$, we have $\mathrm{BO + CO > BC...(iii)}$ and,
In $\Delta$$\mathrm{DOC}$, we have $\mathrm{DO + CO > DC...(iv)}$
Finally, on adding the equations (i),(ii),(iii) and (iv), we get
$\mathrm{AO + OD + AO + BO + BO + CO + DO + CO > AB +AD + BC + DC}$
$\mathrm{= 2(AC + BD) > AB + AD + BC + DC}$ $$\mathrm{= AC + BD > \frac{AB + AD + BC + DC}{2}}$$
The conclusion of the proof is however incorrect and so, somewhere in the proof I think I have blundered.
To whomsoever it may concern, please help me with where did I go wrong.
There is nothing wrong with your arguements. It just proves a completely different result.
Let the sum of the sides be $x$ and the sum of the diagonals be $y$.
The first proof shows that $x>y$, your workings show that $y>\frac{x}{2}.$
Therefore, $$x>y>\frac{x}{2}$$ which essentially bounds (lower and upper) the sum of diagonals.