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

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

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

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

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

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

Cauchy Schwars 

\(M\ge\frac{\left(1+1+1\right)^2}{\left(a+b+c\right)^2}=\frac{9}{\left(a+b+c\right)^2}\ge9\Rightarrow M_{min}=9\Leftrightarrow a=b=c=\frac{1}{3}\)

\(M=\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge\frac{9}{\left(a+b+c\right)^2}\ge9\)

Dau '=' xay ra khi \(a=b=c=\frac{1}{3}\)

Vay \(M_{min}=9\)

các bạn giải rõ ra đc k

ta có \(4=2a^2+\frac{b^2}{4}+\frac{1}{a^2}=a^2+a^2+\frac{b^2}{4}+\frac{1}{a^2}\ge4\sqrt[4]{\frac{a^2.a^2.b^2}{4a^2}}\)

Vậy\(\sqrt[4]{\frac{a^2b^2}{4}}\le1\Leftrightarrow a^2b^2\le4\Leftrightarrow-2\le ab\le2\)

Vậy \(2007\le ab+2009\le2011\)