Khá vắn tắt:

a,

Theo AM-GM có:

\(S_1^2 + S_2^2\ge 2S_1 S_2, S_2^2 +S_3^2\ge 2S_2 S_3,S_1^2 +S_3^2\ge 2S_1S_3\\\to S_1^2 +S_2^2+S_3^2\ge S_1 S_2 +S_2S_3 +S_1S_3\\\to 3(S_1^2 +S_2^2+S_3^2)\ge (S_1+S_2+S_3)^2\\\to 3(S_1^2+S_2^2+S_3^2)\ge 3S^2\\\to S_1^2+S_2^2+S_3^2\ge \dfrac{S^2}{3}\)

Dấu "=" xảy ra khi: \(P\) là trọng tâm \(\Delta ABC\)

b,

Từ \(A\) kẻ \(AV\in B_1C_1(V\in B_1C_1)\)

Qua \(P\) kẻ 1 đường thẳng \(//BC\) cắt \(AB,AC\) tại \(O_3,O_4\)

Qua \(P\) kẻ 1 đường thẳng \(//AB\) cắt \(AC,BC\) tại \(O,O_2\)

Qua \(P\) kẻ 1 đường thẳng \(//AC\) cắt \(AB,BC\) tại \(O_5,O_6\)

Gọi \(N\) là giao của \(BC, AV\to AN\bot BC\)

Gọi \(N_1\) là giao của \(PP_1, BC\to PN_1\bot BC\)

\(VNN_1P_1\) là hình chữ nhật nên \(NV=N_1P_1\)

Thấy: \(\Delta AB_1C_1\backsim \Delta ABC\)

\(\to\dfrac{S_{AB_1C_1}}{S_{ABC}}=\left(\dfrac{AV}{AN}\right)^2\\\to \dfrac{S_{AB_1C_1}}{S_{ABC}}=\left(\dfrac{AN+N_1P_1}{AN}\right)^2\\\to \dfrac{S_{AB_1C_1}}{S_{ABC}}=\left(1+\dfrac{N_1P_1}{AN}\right)^2\\\to S_{AB_1C_1}=\left(1+\dfrac{N_1P_1}{AN}\right)^2 S\\S_{ABC}+S_{BB_1C_1C}=S_{AB_1C_1}\\\to S_{BB_1C_1C}=S_{AB_1C_1}-S_{ABC}=\left(1+\dfrac{N_1P_1}{AN}\right)^2S -S=S\left[\left(1+\dfrac{N_1P_1}{AN}\right)^2-1\right]=S\left(1+\dfrac{2N_1P_1}{AN}+\dfrac{N_1P_1^2}{AN^2}-1\right)=S \left(\dfrac{2N_1P_1}{AN}+\dfrac{N_1P_1^2}{AN^2}\right)\)

Thấy: \(\Delta PO_2O_6\backsim \Delta ABC\)

\(\to \dfrac{PO_2}{AB}=\dfrac{P_1N_1}{AN}\) (Tỉ số đồng dạng bằng tỉ số đường cao)

\(\to \dfrac{BO_3}{AB}=\dfrac{N_1P_1}{AN}\\\to S_{BCC_1B_1}=S\left(\dfrac{2BO_3}{AB}+\dfrac{BO_3^2}{AB^2}\right)\)

Tương tự: \(S_{CAA_2C_2}=S\left(\dfrac{2AO_5}{AB}+\dfrac{AO_5^2}{AB^2}\right)\)

\(S_{ABB_3C_3}=S\left(\dfrac{2O_3O_5}{AB}+\dfrac{O_3O_5^2}{AB^2}\right)\\\to S_{BCC_1B_1}+S_{CAA_2C_2}+S_{ABB_3C_3}=S\left(2+\dfrac{BO_3^2+O_3O_5^2+AO_5^2}{AB^2}\right)\)

Theo AM-GM có:

\(BO_3^2 + O_3O_5^2\ge 2BO_3 . O_3O_5, O_3O_5^2 + AO_5^2\ge 2 O_3O_5 . AO_5, BO_3^2 + AO_5^2\ge 2 BO_3 . AO_5\\\to BO_3^2 + O_3O_5^2 + AO_5^2\ge BO_3 . O_3O_5 + O_3O_5 . AO_5 +AO_5 . BO_3\\\to 3 (BO_3^2 +O_3O_5^2 +AO_5^2)\ge AB^2\\\to S_{BCC_1B_1}+S_{CAA_2C_2}+S_{ABB_3C_3}\ge S\left(2+\dfrac{1}{3}\right)=\dfrac{7}{3}S\)

Dấu "=" xảy ra khi: \(P\) là trọng tâm \(\Delta ABC\)