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\(A=\frac{2}{a^2+b^2}+\frac{35}{ab}+2ab\)
\(=2\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{34}{ab}+\frac{17}{8}ab-\frac{1}{8}ab\)
\(\ge2.\frac{4}{a^2+b^2+2ab}+2\sqrt{\frac{34}{ab}.\frac{17}{8}ab}-\frac{1}{8}.\frac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow A\ge2.\frac{4}{\left(a+b\right)^2}+2.\frac{17}{2}-\frac{1}{8}.\frac{4}{4^2}+17-\frac{1}{2}\)
\(\Leftrightarrow A\ge\frac{1}{2}+17-\frac{1}{2}=17\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=2\)
Chúc bạn học tốt !!!
\(A=\dfrac{2}{a^2+b^2}+\dfrac{35}{ab}+2ab\)
\(=2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\dfrac{34}{ab}+\dfrac{17}{8}ab-\dfrac{1}{8}ab\)
\(\ge2.\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{34}{ab}.\dfrac{17}{8}ab}-\dfrac{1}{8}.\dfrac{\left(a+b\right)^2}{4}\)
\(\Leftrightarrow A\ge2.\dfrac{4}{\left(a+b\right)^2}+2.\dfrac{17}{2}-\dfrac{1}{8}.\dfrac{4^2}{4}\ge2.\dfrac{4}{4^2}+17-\dfrac{1}{2}\)
\(\Leftrightarrow A\ge\dfrac{1}{2}+17-\dfrac{1}{2}=17\)
Dấu "=" <=> 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 \(a=b=c=2\)
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 \(MinP=17\Leftrightarrow a=b=2\)
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
\(A=a^2+b^2+\dfrac{1}{a^2}+\dfrac{1}{b^2}\)
\(A=a^2+\dfrac{1}{16a^2}+b^2+\dfrac{1}{16b^2}+\dfrac{15}{16}\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)\)
\(A\ge2\sqrt{a^2\cdot\dfrac{1}{16a^2}}+2\sqrt{b^2\cdot\dfrac{1}{16b^2}}+\dfrac{15}{16}\cdot2\cdot\sqrt{\dfrac{1}{a^2b^2}}\)
\(A\ge1+\dfrac{15}{8ab}\ge1+\dfrac{15}{2\left(a+b\right)^2}\ge\dfrac{17}{2}\)
"="<=>x=y=0,5
\(P=\frac{2}{a^2+b^2}+\frac{2}{2ab}+\frac{34}{ab}+\frac{17ab}{8}-\frac{ab}{8}\)
\(P=2\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{34}{ab}+\frac{17ab}{8}-\frac{ab}{8}\)
\(P\ge2\cdot\frac{4}{a^2+b^2+2ab}+2\sqrt{\frac{34}{ab}\cdot\frac{17ab}{8}}-\frac{\frac{\left(a+b\right)^2}{4}}{8}\)
( do \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y};x+y\ge2\sqrt{xy};ab\le\frac{\left(a+b\right)^2}{4}\))
\(\Rightarrow P\ge\frac{8}{\left(a+b\right)^2}+2\sqrt{\frac{289}{4}}-\frac{\frac{4^2}{4}}{8}\)
\(\Rightarrow P\ge\frac{8}{16}+17-\frac{1}{2}=17\)
\(P=17\Leftrightarrow\left\{{}\begin{matrix}a^2+b^2=2ab\\\frac{34}{ab}=\frac{17ab}{8}\\a=b\\a+b=4\end{matrix}\right.\Leftrightarrow a=b=2\)
Vậy Min P = 17 \(\Leftrightarrow a=b=2\)
Dự đoán dấu "=" khi \(a=b=c \Rightarrow P=28\)
Ta sẽ chứng minh \(P=28\) là GTNN
Thật vậy ta có: \(P=\dfrac{ab+bc+ca}{a^2+b^2+c^2}-1+\dfrac{\left(a+b+c\right)^3}{abc}-27\ge0\)
\(\Leftrightarrow\dfrac{ab+bc+ca-\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}+\dfrac{\left(a+b+c\right)^3-27abc}{abc}\ge0\)
\(\Leftrightarrow\dfrac{\left(a+b+c\right)^3-27abc}{abc}-\dfrac{2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)}{2\left(a^2+b^2+c^2\right)}\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(\dfrac{\dfrac{a+b+7c}{2}\cdot\left(a-b\right)^2}{abc}-\dfrac{\left(a-b\right)^2}{2\left(a^2+b^2+c^2\right)}\right)\ge0\)
\(\LeftrightarrowΣ_{cyc}\left(\left(a-b\right)^2\left(\dfrac{a+b+7c}{2abc}-\dfrac{1}{2\left(a^2+b^2+c^2\right)}\right)\right)\ge0\) *Đúng*
Vậy ...
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 \(A_{\min}=17\Leftrightarrow a=b=2\)