Áp dụng BĐT BSC và Cosi:

\(\dfrac{1}{a^2+b^2}+\dfrac{2}{ab}+4ab=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{4ab}+4ab+\dfrac{5}{4ab}\)

\(\ge\dfrac{4}{a^2+b^2+2ab}+2\sqrt{\dfrac{1}{4ab}.4ab}+\dfrac{5}{\left(a+b\right)^2}\)

\(=\dfrac{4}{\left(a+b\right)^2}+2+\dfrac{5}{\left(a+b\right)^2}\ge4+2+5=11\)

\(min=11\Leftrightarrow a=b=\dfrac{1}{2}\)

vì a;b>0\(\Rightarrow a+b>=2\sqrt{ab}\Rightarrow1>=2\sqrt{ab}\Rightarrow\frac{1}{2}>=\sqrt{ab}\Rightarrow\frac{1}{4}>=ab\)(bđt cosi)

dấu = xảy ra khi a=b=\(\frac{1}{2}\)

\(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2=1+\frac{2}{a}+\frac{1}{a^2}+1+\frac{2}{b}+\frac{1}{b^2}\)

\(=2+\left(\frac{2}{a}+\frac{2}{b}\right)+\left(\frac{1}{a^2}+\frac{1}{b^2}\right)>=2+2\sqrt{\frac{2}{a}\cdot\frac{2}{b}}+2\cdot\sqrt{\frac{1}{a^2}\cdot\frac{1}{b^2}}\)(bđt cosi )

dấu = xảy ra khi \(\frac{2}{a}=\frac{2}{b}\Rightarrow a=b=\frac{1}{2};\frac{1}{a^2}=\frac{1}{b^2}\Rightarrow a=b=\frac{1}{2}\)\(\Rightarrow\)dấu = xảy ra khi \(a=b=\frac{1}{2}\)

\(=2+\frac{4}{\sqrt{ab}}+\frac{2}{\sqrt{a^2b^2}}=2+\frac{4}{\sqrt{ab}}+\frac{2}{ab}>=2+\frac{4}{\frac{1}{2}}+\frac{2}{\frac{1}{4}}=2+8+8=18\)

\(\Rightarrow M>=18\Rightarrow\)min M là 18

vậy min M là 18 khi a=b=\(\frac{1}{2}\)

Áp dụng bđt Cauchy-Schwarz dạng Engel ta có :

\(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2=\frac{\left(1+\frac{1}{a}\right)^2}{1}+\frac{\left(1+\frac{1}{b}\right)^2}{1}\ge\frac{\left(1+\frac{1}{a}+1+\frac{1}{b}\right)^2}{2}=\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)}{2}\)(1)

Lại có \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}=4\)(2) 

Từ (1) và (2) => \(M=\left(1+\frac{1}{a}\right)^2+\left(1+\frac{1}{b}\right)^2\ge\frac{\left(2+\frac{1}{a}+\frac{1}{b}\right)^2}{2}\ge\frac{\left(2+4\right)^2}{2}=18\)

Đẳng thức xảy ra khi a = b = 1/2

Vậy MinM = 18, đạt được khi a = b = 1/2

\(P=\left(\frac{a}{b}+\frac{b}{a}\right)^2-\left(\frac{a}{b}+\frac{b}{a}\right)-1\)

Đặt \(t=\frac{a}{b}+\frac{b}{a}\Rightarrow t\in(-\infty;-2]\cup[2;+\infty)\)

\(f\left(t\right)=t^2-t-1\)

\(-\frac{b}{2a}=\frac{1}{2}\notin(-\infty;-2]\cup[2;+\infty)\)

\(f\left(-2\right)=5\) ; \(f\left(2\right)=1\)

\(\Rightarrow P_{min}=1\) khi \(t=2\Leftrightarrow\frac{a}{b}+\frac{b}{a}=2\Leftrightarrow a=b\)

Áp dụng BĐT AM-GM ta có:

\(a+b\ge2\sqrt{ab}\Rightarrow1\ge2\sqrt{ab}\Rightarrow\dfrac{1}{2}\ge\sqrt{ab}\Rightarrow\dfrac{1}{4}\ge ab\)

Lại có theo AM-GM ta có:

\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)\(\Rightarrow\dfrac{3}{a^2+b^2}\ge\dfrac{3}{2ab}\)

\(\Rightarrow A\ge\dfrac{3}{2ab}+\dfrac{2}{ab}\ge\dfrac{3}{2\cdot\dfrac{1}{4}}+\dfrac{2}{\dfrac{1}{4}}=14\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}a+b=2\sqrt{ab}\\a+b=1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=b\\a+b=1\end{matrix}\right.\)\(\Rightarrow a=b=\dfrac{1}{2}\)

Vậy \(A_{Min}=14\) khi \(a=b=\dfrac{1}{2}\)

\(A=\dfrac{3}{a^2+b^2}+\dfrac{3}{2ab}+\dfrac{1}{2ab}\ge\dfrac{12}{\left(a+b\right)^2}+\dfrac{2}{\left(a+b\right)^2}=14\)

dùng cô si, chiều mình làm cho bạn xem, gợi ý trước, dùng cô si áp dụng lên cái a+b=1 kia kìa, rồi giải ra, thế thôi, còn ko làm được, chièu mình làm cho, giờ đang bận

Lời gải:

Áp dụng BĐT Cauchy Schwarz và BĐT AM-GM:

$M=\frac{1}{2ab}+\frac{1}{2ab}+\frac{1}{a^2+ab}+\frac{1}{b^2+ab}+\frac{1}{a^2+b^2}$

$\geq \frac{(1+1+1+1+1)^2}{2ab+2ab+a^2+ab+b^2+ab+a^2+b^2}=\frac{25}{2a^2+2b^2+6ab}$

$=\frac{25}{2(a^2+b^2+2ab)+2ab}$

$=\frac{25}{2(a+b)^2+2ab}=\frac{25}{2+2ab}\geq \frac{25}{2+2.\frac{(a+b)^2}{4}}=\frac{25}{2+\frac{2}{4}}=10$

Vậy  $M_{\min}=10$. Giá trị này đạt tại $a=b=\frac{1}{2}$

Có: \(a^2+b^2\ge2ab\Rightarrow a^2+b^2\ge2\)
\(\Rightarrow\left(a+b+1\right)\left(a^2+b^2\right)\ge2\left(a+b+1\right)\)
\(\Rightarrow Q\ge2\left(a+b\right)+\frac{8}{a+b}+2\)
Mà: \(2\left(a+b\right)+\frac{8}{a+b}\ge2\sqrt{2\left(a+b\right).\frac{8}{a+b}}=8\)
\(\Rightarrow Q\ge10\)
Dấu "=" xảy ra <=> a=b=1