Ta có:

\((a+b)^2 \leq 16 \Rightarrow a^2+b^2 \leq 16-2ab \)

\((a+b)^2 \geq 4ab \Rightarrow ab \leq 4 \)

Suy ra \(P\ge\dfrac{1}{8-ab}+\dfrac{35}{ab}+2ab\)

\(=\dfrac{1}{8-ab}+\dfrac{8-ab}{16}+\dfrac{33ab}{16}+\dfrac{33}{ab}+2ab-\dfrac{1}{2}\)

\(\ge\dfrac{2\cdot1}{4}+\dfrac{2\cdot33}{4}+\dfrac{2}{4}-\dfrac{1}{2}=17\)

Dấu "=" xảy ra khi \(a=b=2\)

ta có

\(\left(a+b\right)^2\ge4ab\Rightarrow ab\le4\)\(P=2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\left(\dfrac{32}{ab}+2ab\right)+\dfrac{2}{ab}\ge2\dfrac{4}{\left(a+b\right)^2}+2\sqrt{\dfrac{32}{ab}.2ab}+\dfrac{2}{4}=\dfrac{8}{16}+2.8+\dfrac{1}{2}=17.\)

P min=17 khi a=b=2

\(P=2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\left(\dfrac{32}{ab}+2ab\right)\) đề bài 35 =32 mới đúng nhỉ ( bạn ơi)

\(A=\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\left(ab+\dfrac{16}{ab}\right)+\dfrac{17}{2ab}\)

\(A\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{16ab}{ab}}+\dfrac{17}{\dfrac{2\left(a+b\right)^2}{4}}\)

\(A\ge\dfrac{4}{\left(a+b\right)^2}+8+\dfrac{34}{\left(a+b\right)^2}\ge\dfrac{4}{4^2}+8+\dfrac{34}{4^2}=\dfrac{83}{8}\)

Dấu "=" xảy ra khi \(a=b=2\)

Lời giải:

Ta dự toán cực trị xảy ra tại \(a=b=2\). Công việc còn lại là phân tích hợp lý.

Áp dụng BĐT Bunhiacopxky:

\(\left(\frac{2}{a^2+b^2}+\frac{2}{2ab}\right)(a^2+b^2+2ab)\geq (\sqrt{2}+\sqrt{2})^2\)

\(\Leftrightarrow \frac{2}{a^2+b^2}+\frac{1}{ab}\geq \frac{8}{a^2+b^2+2ab}=\frac{8}{(a+b)^2}\)

Mà \(a+b\lè 4\Rightarrow \frac{2}{a^2+b^2}+\frac{1}{ab} \geq \frac{8}{(a+b)^2}\geq \frac{8}{4^2}=\frac{1}{2}(1)\)

Áp dụng BĐT AM-GM:

\(\frac{32}{ab}+2ab\geq 2\sqrt{32.2}=16(2)\)

Tiếp tục AM-GM: \(4\geq a+b\geq 2\sqrt{ab}\Rightarrow ab\leq 4\)

\(\Rightarrow \frac{2}{ab}\geq \frac{2}{4}=\frac{1}{2}(3)\)

Lấy \((1)+(2)+(3)\Rightarrow A\geq \frac{1}{2}+16+\frac{1}{2}=17\)

Vậy \(A_{\min}=17\Leftrightarrow a=b=2\)

\(a\ge2b\Rightarrow\dfrac{a}{b}\ge2\)

\(P=2\left(\dfrac{a}{b}\right)+\left(\dfrac{b}{a}\right)-2=\dfrac{a}{4b}+\dfrac{b}{a}+\dfrac{7}{4}\left(\dfrac{a}{b}\right)-2\ge2\sqrt{\dfrac{ab}{4ab}}+\dfrac{7}{4}.2-2=\dfrac{5}{2}\)

\(P_{min}=\dfrac{5}{2}\) khi \(a=2b\)

Ta có : \(4\ge a+b\ge2\sqrt{ab}\Rightarrow ab\le4\)

Áp dụng bất đẳng thức \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)(bạn có thể chứng minh bằng biến đổi tương đương)

Ta có :\(P=\frac{2}{a^2+b^2}+\frac{35}{ab}+2ab=\left(\frac{2}{a^2+b^2}+\frac{1}{ab}\right)+\left(\frac{32}{ab}+2ab\right)+\frac{2}{ab}=2\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\left(\frac{32}{ab}+2ab\right)+\frac{2}{ab}\ge\frac{2.4}{\left(a+b\right)^2}+2\sqrt{\frac{32}{ab}.2ab}+\frac{2}{ab}\ge\frac{8}{4^2}+2.8+\frac{2}{4}=17\)Dấu đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}a=b\\a^2b^2=16\\0< a+b\le4\end{cases}\Leftrightarrow}a=b=2\)

Vậy \(MinP=17\Leftrightarrow a=b=2\)

\(A=\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\\ =\dfrac{2}{a^2+b^2}+\dfrac{2}{2ab}+\dfrac{34}{ab}+\dfrac{17ab}{8}-\dfrac{ab}{8}\\ =2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+17\left(\dfrac{2}{ab}+\dfrac{ab}{8}\right)-\dfrac{ab}{8}\\ \overset{AM-GM}{\ge}2\cdot\dfrac{1}{a^2+b^2+2ab}+17\sqrt{\dfrac{2}{ab}\cdot\dfrac{ab}{8}}-\dfrac{\left(a+b\right)^2}{4\cdot8}\\ =\dfrac{2}{\left(a+b\right)^2}+\dfrac{17}{2}-\dfrac{\left(a+b\right)^2}{32}\\ \ge\dfrac{2}{4^2}+\dfrac{17}{2}-\dfrac{4^2}{32}=\dfrac{65}{8}\)

Dấu "=" xảy ra khi : \(\left\{{}\begin{matrix}\dfrac{2}{ab}=\dfrac{ab}{8}\\a^2+b^2=2ab\\a=b\\a+b=4\end{matrix}\right.\Leftrightarrow a=b=2\)

Vậy \(A_{Min}=\dfrac{65}{8}\) khi \(a=b=2\)

\(\ge2\cdot\dfrac{4}{a^2+b^2+2ab}+17\cdot2\sqrt{\dfrac{2}{ab}+\dfrac{ab}{8}}-\dfrac{\left(a+b\right)^2}{4\cdot8}\\ =\dfrac{8}{\left(a+b\right)^2}+17-\dfrac{\left(a+b\right)^2}{32}\\ \ge\dfrac{8}{4^2}+17-\dfrac{4^2}{32}=17\)

Vậy \(A_{Min}=17\) khi \(a=b=c=2\)

\(A=\frac{2}{a^2+b^2}+\frac{35}{ab}+2ab\)

\(=2\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{34}{ab}+\frac{17}{8}ab-\frac{1}{8}ab\)

\(\ge2.\frac{4}{a^2+b^2+2ab}+2\sqrt{\frac{34}{ab}.\frac{17}{8}ab}-\frac{1}{8}.\frac{\left(a+b\right)^2}{4}\)

\(\Leftrightarrow A\ge2.\frac{4}{\left(a+b\right)^2}+2.\frac{17}{2}-\frac{1}{8}.\frac{4}{4^2}+17-\frac{1}{2}\)

\(\Leftrightarrow A\ge\frac{1}{2}+17-\frac{1}{2}=17\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=2\)

Chúc bạn học tốt !!!