a=b=c=2 thay vào ra min cái này là tay tui tự gõ ra a=b=c=2 chả có bước nào. còn chi tiết sau nhớ nhắc tui làm :D

Áp dụng BĐT Mincopxki và AM-GM có:

\(T=\sqrt{a^2+\frac{1}{b^2}}+\sqrt{b^2+\frac{1}{c^2}}+\sqrt{c^2+\frac{1}{a^2}}\)

\(\ge\sqrt{\left(a+b+c\right)^2+\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}\)

\(\ge\sqrt{\left(a+b+c\right)^2+\frac{81}{\left(a+b+c\right)^2}}\)

\(=\sqrt{\frac{81}{\left(a+b+c\right)^2}+\frac{\left(a+b+c\right)^2}{16}+\frac{15\left(a+b+c\right)^2}{16}}\)

\(=\sqrt{2\sqrt{\frac{81}{\left(a+b+c\right)^2}\cdot\frac{\left(a+b+c\right)^2}{16}}+\frac{15\cdot6^2}{16}}\)

\(=\sqrt{2\sqrt{\frac{81}{16}}+\frac{15\cdot6^2}{16}}=\frac{3\sqrt{17}}{2}\)

Khi \(a=b=c=2\)

\(M=\left(a^2+\frac{1}{16a^2}\right)+\left(b^2+\frac{1}{16b^2}\right)+\frac{15}{16}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)\)

\(\ge2\sqrt{\frac{a^2}{16a^2}}+2\sqrt{\frac{b^2}{16b^2}}+\frac{15\left(\frac{1}{a}+\frac{1}{b}\right)^2}{32}\ge1+\frac{\frac{240}{\left(a+b\right)^2}}{32}\ge\frac{17}{2}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=\frac{1}{2}\)

Ta thấy: \(a+b\le1\Leftrightarrow\hept{\begin{cases}a\le1-b\\b\le1-a\end{cases}}\Leftrightarrow\hept{\begin{cases}1+a\le2-b\\1+b\le2-a\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}\frac{a}{1+b}\ge\frac{a}{2-a}\\\frac{b}{1+a}\ge\frac{b}{2-b}\end{cases}}\Rightarrow\frac{a}{1+b}+\frac{b}{1+a}\ge\frac{a}{2-a}+\frac{b}{2-b}\)

\(\Rightarrow S=\frac{a}{1+b}+\frac{b}{1+a}+\frac{1}{a+b}\ge\frac{a}{2-a}+\frac{b}{2-b}+\frac{1}{a+b}\)

\(=\frac{2}{2-a}-1+\frac{2}{2-b}-1+\frac{1}{a+b}=\frac{2}{2-a}+\frac{2}{2-b}+\frac{1}{a+b}-2\)

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

Áp dụng bất đẳng thức sau: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}\)

\(\Rightarrow\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2\left(a+b\right)}\ge\frac{9}{4-\left(a+b\right)+2\left(a+b\right)}=\frac{9}{4+a+b}\)

Lại có: \(a+b\le1\Rightarrow4+a+b\le5\Rightarrow\frac{9}{4+a+b}\ge\frac{9}{5}\)

\(\Rightarrow\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2\left(a+b\right)}\ge\frac{9}{5}\Leftrightarrow2\left(\frac{1}{2-a}+\frac{1}{2-b}+\frac{1}{2\left(a+b\right)}-1\right)\ge\frac{8}{5}\)

\(\Rightarrow S\ge\frac{8}{5}.\)

Vậy \(Min_S=\frac{8}{5}.\)Dấu "=" xảy ra khi \(a=b=\frac{2}{5}.\)

Ta có

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

\(\ge2+2+a+b+\frac{4}{a+b}\)

\(=4+a+b+\frac{2}{a+b}+\frac{2}{a+b}\)

 \(\ge4+2\sqrt{\left(a+b\right).\frac{2}{\left(a+b\right)}}+\frac{2}{\sqrt{2\left(a^2+b^2\right)}}\)

\(=4+2\sqrt{2}+\sqrt{2}=4+3\sqrt{2}\)

Đặt \(a-1=x>0,b-1=y>0\), ta có

\(A=\frac{\left(x+1\right)^2}{x}+\frac{\left(y+1^2\right)}{y}=\frac{x^2+2x+1}{x}+\frac{y^2+2y+1}{y}\)

\(=\left(x+\frac{1}{x}\right)+\left(y+\frac{1}{y}\right)+4\)

Với \(x>0,y>0\)ta có \(x+\frac{1}{x}\ge2,y+\frac{1}{y}\ge2\)nên \(A\ge8\)

\(Min_A=8\Leftrightarrow x=y=1\Leftrightarrow a=b=2\)

P/s tham khảo nha

Sử dụng \(AM-GM\)ta có :

\(\frac{a^2}{a-1}+4\left(a-1\right)\ge2\sqrt{\left(2a\right)^2}=4a\)

Tương tự : \(\frac{b^2}{b-1}+4\left(b-1\right)\ge4b\)

Cộng theo vế : \(A+4\left(a+b\right)-8\ge4\left(a+b\right)\)

\(< =>A\ge8\)

Dấu = xảy ra \(< =>a=b=2\)