Dễ dàng để ý thì $\dfrac{OK}{AK},\dfrac{OP}{CP}$ không liên quan gì đến nhau nên nghĩ cách thay các tỉ số bằng tỉ số mới rồi dùng Cauchy.

$\Delta OKD\backsim\Delta NCD$ (g.g)

$\to \dfrac{OK}{CN}=\dfrac{OD}{DN}\\\to OK.DN=CN.OD\\\to OK=\dfrac{CN.OD}{DN}$

$\Delta AKD\backsim\Delta NAD$ (g.g)

$\to \dfrac{AK}{NA}=\dfrac{AD}{ND}\\\to AK.ND=NA.AD\\\to AK=\dfrac{AN.AD}{ND}\\\to \dfrac{OK}{AK}=\dfrac{\dfrac{CN.OD}{DN}}{\dfrac{AN.AD}{DN}}=\dfrac{CN.OD}{AN.AD}=\dfrac{CN.R}{AN.R\sqrt{2}}$

(Dễ dàng thấy $\Delta AOD$ vuông cân tại $O$ nên theo Pytago $AD=R\sqrt{2}$)

$\to \dfrac{OK}{AK}=\dfrac{CN}{AN\sqrt{2}}$

Tương tự: $\dfrac{OP}{CP}=\dfrac{AN}{CN\sqrt{2}}$

Như vậy: $\dfrac{OK}{AK}+\dfrac{OP}{CP}=\dfrac{1}{\sqrt{2}} \left(\dfrac{CN}{AN}+\dfrac{AN}{CN}\right)$

Theo Cauchy: $\dfrac{CN}{AN}+\dfrac{AN}{CN}\ge 2\sqrt{\dfrac{CN}{AN}.\dfrac{AN}{CN}}=2$

$\to \dfrac{OK}{AK}+\dfrac{OP}{CP}\ge \dfrac{2}{\sqrt{2}}=\sqrt{2}$

Dấu "=" xảy ra khi:

$\dfrac{OK}{AK}=\dfrac{OP}{CP},\dfrac{OK}{AK}+\dfrac{OP}{CP}=\sqrt{2}\\\to \dfrac{OK}{AK}=\dfrac{1}{\sqrt{2}}\\\to \dfrac{OK}{OA}=\dfrac{1}{\sqrt{2}+1}\\\to OK.(\sqrt{2}+1)=OA=R\\\to OK=\dfrac{R}{\sqrt{2}+1}$