Algebraic equation to represent a triangle.

The lengths of the sides of a triangle are $p, q, r$ respectively. If $p^2 + q^2 + r^2= pq + qr + pr$, then this triangle is

(A) equilateral triangle

(B) isosceles triangle

(C) right angled triangle

(D) obtuse triangle

My Attempt: \begin{align} &p^2 + q^2 + r^2= pq + qr + pr \\ &\implies (p^2-pq) + (q^2-qr) +( r^2-pr)=0 \\ &\implies p(p-q)+q(q-r)+r(r-p)=0, \end{align} but I'm unable to connect this information to the given options. Please help me.

Since $p^2+q^2+r^2 \geq pq+qr+rp$ for $p,q,r \in R^{+}$

But given that $p^2+q^2+r^2 = pq+qr+rp$.

Equality holds when $p=q=r$.

Hence equilateral triangle.

NOTE:

$(p-q)^2\geq 0$ , $p^2+q^2\geq 2pq$

$(q-r)^2\geq 0$ , $q^2+r^2\geq 2qr$

$(r-p)^2\geq 0$ , $r^2+p^2\geq 2rp$

then

$p^2+q^2+r^2 \geq pq+qr+rp$ for $p,q,r \in R^{+}$

Let me clarify Person Average's suggestion: \begin{align} &2(p^2+q^2+r^2)=2(pq+qr+rp) \\ &\implies (p^2-2pq+q^2)+(q^2-2qr+r^2)+(r^2-2rp+p^2)=0 \\ &\implies (p-q)^2+(q-r)^2+(r-p)^2=0 \\ &\implies p=q=r. \end{align}

What you need is just multiply $2$ for left hand and right hand , then you can find that it is $(3\times \text{square}) = 0$ and they must be all zero . so it is a equilateral triangle .