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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 !!!
\(a+b\ge2\sqrt{ab}\Leftrightarrow2\sqrt{ab}\le4\Leftrightarrow ab\le4\)
\(P=\left(\dfrac{2}{a^2+b^2}+\dfrac{1}{ab}\right)+\dfrac{2}{ab}+2ab+\dfrac{32}{ab}\\ \Leftrightarrow P=2\left(\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\right)+\dfrac{2}{ab}+2ab+\dfrac{32}{ab}\\ \Leftrightarrow P\ge2\cdot\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{32}{ab}\cdot2ab}+\dfrac{2}{4}\\ \Leftrightarrow P\ge\dfrac{8}{\left(a+b\right)^2}+2\sqrt{64}+\dfrac{1}{2}\\ \Leftrightarrow P\ge\dfrac{8}{16}+16+\dfrac{1}{2}=17\)
Dấu \("="\Leftrightarrow a=b=2\)
\(P=\frac{2}{a^2+b^2}+\frac{2}{2ab}+\frac{34}{ab}+\frac{17ab}{8}-\frac{ab}{8}\)
\(P=2\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\frac{34}{ab}+\frac{17ab}{8}-\frac{ab}{8}\)
\(P\ge2\cdot\frac{4}{a^2+b^2+2ab}+2\sqrt{\frac{34}{ab}\cdot\frac{17ab}{8}}-\frac{\frac{\left(a+b\right)^2}{4}}{8}\)
( 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 \(\Leftrightarrow a=b=2\)
Áp dụng BĐT \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) và BĐT AM-GM ta có:
\(P=\frac{2}{a^2+b^2}+\frac{2}{2ab}+\frac{32}{ab}+2ab+\frac{2}{ab}\)
\(\ge\frac{2.4}{a^2+b^2+2ab}+2\sqrt{\frac{32}{ab}.2ab}+\frac{2}{ab}\)
\(\ge\frac{8}{\left(a+b\right)^2}+2.\sqrt{64}+\frac{2}{\frac{\left(a+b\right)^2}{4}}\)
\(\ge\frac{8}{4^2}+2.8+\frac{8}{\left(a+b\right)^2}\ge\frac{1}{2}+16+\frac{8}{4^2}=\frac{1}{2}+16+\frac{1}{2}=17\)
Nên GTNN của P là 17 đạt được khi a=b=2
Áp dụng bất đẳng thức Cosi ta có :
\(4\ge a+b\ge2\sqrt{ab}\Leftrightarrow\sqrt{ab}\le2\Leftrightarrow ab\le4\)
Ta có bất đẳng thức \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)
(Nhân chéo để chứng minh )
Áp dụng :
\(S=\frac{1}{a^2+b^2}+\frac{25}{ab}+ab=\frac{1}{a^2+b^2}+\frac{1}{2ab}+\frac{49}{2ab}+ab\)
\(=\frac{1}{a^2+b^2}+\frac{1}{2ab}+ab+\frac{16}{ab}+\frac{17}{2ab}\)
\(\ge\frac{4}{a^2+b^2+2ab}+2\sqrt{ab.\frac{16}{ab}}+\frac{17}{2ab}\)
\(\ge\frac{4}{\left(a+b\right)^2}+8+\frac{17}{2.4}=\frac{1}{4}+8+\frac{17}{8}=\frac{83}{8}\)
Dấu " = " xảy ra \(\Leftrightarrow a=b=2\)
\(S=\frac{1}{a^2+b^2}+\frac{25}{ab}+ab\)
\(=\left(\frac{1}{a^2+b^2}+\frac{1}{2ab}\right)+\left(ab+\frac{16}{ab}\right)+\frac{17}{2ab}\)
\(\ge\frac{4}{\left(a+b\right)^2}+2\sqrt{ab\cdot\frac{16}{ab}}+\frac{17}{\frac{\left(a+b\right)^2}{2}}\)
\(\ge\frac{4}{4^2}+8+\frac{17}{\frac{4^2}{2}}=\frac{83}{8}\)
Dấu "=" xảy râ khi x = y = 2
Ta có \(a+b\ge2\sqrt{ab}\)=> \(ab\le4\)
\(\frac{1}{a^2+b^2}+\frac{1}{2ab}\ge\frac{4}{\left(a+b\right)^2}\ge\frac{1}{4}\)
\(\frac{16}{ab}+ab\ge8\)
\(\frac{17}{2ab}\ge\frac{17}{8}\)
=> \(S\ge8+\frac{17}{8}+\frac{1}{4}=\frac{83}{8}\)
Vậy MinS=83/8 khi 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
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 \(MinP=17\Leftrightarrow a=b=2\)