deletions | additions       diff --git a/untitled.tex b/untitled.tex  index 24b1040..ed12509 100644  --- a/untitled.tex  +++ b/untitled.tex 
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Thus, $$(x+\frac{1}{2}c)^2 + (y-\frac{1}{2}d)^2 = -b + \frac{1}{4}|a^2|$$  Notice the left hand side $$(x+\frac{1}{2}c)^2 + (y-\frac{1}{2}d)^2 \geq 0$$  Then the right hand side $$ -b + \frac{1}{4}|a^2|\geq 0.$$  Thus, $\frac{1}{4}|a^2|)\geq $\frac{1}{4}|a^2|\geq  b$\\ Hence, $|a^2|\geq 4b$  \end{proof}