Existence of a quadrilateral with side lengths $a,b,c,d$ where $a+b+c>d$ and $d$ is largest length I tried to prove triangle inequality, given side lengths $a, b, c$ if $a + b \gt c$, then a triangle always exists, by placing the side length $c$ on the $x$-axis and proving that intersection points of two circles of radius $b$ and $c$ always exists.
Now, I wanted to do the same for any polygon with number of sides greater than $3$. I tried to prove for quadrilateral by using same method. But I can't progress.
Any idea on how prove if sides are $a, b, c, d$, where $d$ is the largest side and $a + b + c \gt d$, then quadrilateral exists.
 A: Assume $a < b < c < d$.
Suppose that WLOG, $a+b > d$. Then , construct a triangle with lengths $a,b,d$, and at the vertex joining $b$ and $d$, make a separation and attach the length $c$ to make a quadrilateral. Similarly for the other pairwise sums. Hence, we will assume that each pairwise sum is less than $d$.
Let $\epsilon = a+b+c-d > 0$.
A triangle exists with length $a,b,$ and $ d-c + \frac{\epsilon}{2}$ , as $d-c + \frac{\epsilon}2 < d-c + \epsilon = a+b$, and $a < b$, and $b+c < d \implies b < d-c + \frac{\epsilon}{2}$.
Furthermore, a triangle exists with lengths $d-c + \frac{\epsilon}{2} , d, c$ as $c+d-c+\frac{\epsilon}{2} > d$ , $c< d$, and if $d-c + \frac{\epsilon}{2} >  d +c \implies c < \frac{\epsilon}{4} \implies a+b+c < \epsilon \implies a+b+c < a+b+c-d$, which is a contradiction.
Now, you have two triangles having the side $d-c+\frac{\epsilon}2$ common. Glue them together to get a quadrilateral with sides $a,b,c,d$.


A: Unlike the triangle case, a quadrilateral is not uniquely defined by the lengths of its sides. Because of that, the line of proof that worked in the triangle case will not (directly) work for a quadrilateral. And, it's not clear from the context what level of rigor you are aiming for.
As a possible approach, construct the triangle with sides $a+b,c,d$ which exists per the triangle case that you already proved. Then pick the point on side $a+b$ that divides it into lengths $a,b$ and shift it slightly while keeping $c,d$ constant (as in a rigid articulated polygon), which will form the 4th vertex of a quadrilateral proper.
