Theo t/c tiếp tuyến thì \(\left\{{}\begin{matrix}BP=PE\\DQ=QE\end{matrix}\right.\)
\(\Rightarrow BP+DQ=PE+QE=PQ\)
\(\Rightarrow\left(BP+PC\right)+\left(DQ+QC\right)=PQ+PC+QC\)
\(\Rightarrow BC+DC=PQ+PC+QC\)
\(\Rightarrow4=PQ+PC+QC\le PQ+\sqrt{2\left(PC^2+QC^2\right)}\)
(sử dụng BĐT \(\left(a+b\right)^2\le2\left(a^2+b^2\right)\))
\(\Rightarrow4\le PQ+\sqrt{2PQ^2}\)
\(\Rightarrow4\le PQ\left(1+\sqrt{2}\right)\)
\(\Rightarrow PQ\ge\dfrac{4}{1+\sqrt{2}}\)
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