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

S = a+b+c + (1/a + 1/b + 1/c)

   >= (a+b+c) + 9/a+b+c

    = [ (a+b+c) + 9/4.(a+b+c) ] + 27/4.(a+b+c)

   >= \(2\sqrt{\left(a+b+c\right).\frac{9}{4.\left(a+b+c\right)}}\)   +    27/(4.3/2)

     = 3 + 9/2

     = 15/2

Dấu "=" xảy ra <=> a=b=c=1/2

Vậy ......

Tk mk nha

bài này còn có thể theo phương pháp chọn điểm rơi trong bài toán cực trị, bạn thử tìm hiểu nhé!!!!

Đầu tiên ta chứng minh bđt:\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\Leftrightarrow\dfrac{a+b}{ab}\ge\dfrac{4}{a+b}\)

\(\Leftrightarrow\left(a+b\right)^2\ge4ab\)

\(\Leftrightarrow a^2+2ab+b^2-4ab\ge0\)

\(\Leftrightarrow\left(a-b\right)^2\ge0\)(luôn đúng)

Áp dụng \(\Rightarrow P=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}\ge\dfrac{4}{a^2+b^2+2ab}=\dfrac{4}{\left(a+b\right)^2}\ge\dfrac{4}{4^2}=\dfrac{1}{4}\)

\(\Rightarrow MINP=\dfrac{1}{4}\Leftrightarrow a=b=2\)

+)\(\frac{3}{4}\ge a^2+b^2+c^2\ge3\sqrt[3]{a^2b^2c^2}\Leftrightarrow\frac{1}{8}\ge abc\)

+) \(P=8abc+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(32abc+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)-24abc\)

\(\ge4\sqrt[4]{\frac{32}{abc}}-24abc\ge4\sqrt[4]{\frac{32}{\frac{1}{8}}}-3=16-3=13\)

Dấu = xảy ra khi \(a=b=c=\frac{1}{2}\)

\(S=\left(a^2+b^2+c^2+\frac{1}{8a}+\frac{1}{8b}+\frac{1}{8c}+\frac{1}{8a}+\frac{1}{8b}+\frac{1}{8c}\right)+\frac{3}{4a}+\frac{3}{4b}+\frac{3}{4c}\)

\(\ge9\sqrt[9]{a^2b^2c^2.\frac{1}{8a}.\frac{1}{8b}.\frac{1}{8c}.\frac{1}{8a}.\frac{1}{8b}.\frac{1}{8c}}+\frac{3}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\ge\frac{9}{4}+9.\frac{1}{\sqrt[3]{abc}}\ge\frac{9}{4}+\frac{9}{4}.\frac{1}{\frac{a+b+c}{3}}\ge\frac{9}{4}+\frac{9}{4}.2=\frac{27}{4}\)

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

Vậy \(Min_S=\frac{27}{4}\)

Ta thấy:

\(\left(a^2+2bc\right)+\left(b^2+2ac\right)+\left(c^2+2ab\right)=\left(a+b+c\right)^2\le1\)

Áp dụng BĐT AM-GM ta có:

\(P\ge\left[\left(a^2+2bc\right)+\left(b^2+2ac\right)+\left(c^2+2ab\right)\right]\left(\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\right)\)

\(\ge3\sqrt[3]{\left(a^2+2bc\right)\left(b^2+2ac\right)\left(c^2+2ab\right)}\cdot3\sqrt[3]{\frac{1}{a^2+2bc}\cdot\frac{1}{b^2+2ac}\cdot\frac{1}{c^2+2ab}}=9\)

Dấu "="xảy ra khi \(\left\{\begin{matrix}a+b+c=1\\a^2+2bc=b^2+2ac=c^2+2ab\end{matrix}\right.\)\(\Rightarrow a=b=c=\frac{1}{3}\)

Vậy \(Min_P=9\) khi \(a=b=c=\frac{1}{3}\)

B1 

Ta có

\(A=\frac{a^2}{24}+\frac{9}{a}+\frac{9}{a}+\frac{23a^2}{24}\ge3\sqrt[3]{\frac{a^2}{24}.\frac{9}{a}.\frac{9}{a}+\frac{23a^2}{24}}\ge\frac{9}{2}+\frac{23.36}{24}\ge39\)

Dấu "=" xảy ra <=> a=6

Vậy Min A = 39 <=> a=6

 \(A=a^2+\frac{18}{a}=a^2+\frac{216}{a}+\frac{216}{a}-\frac{414}{a}\ge3\sqrt[3]{a^2.\frac{216}{a}.\frac{216}{a}}-69=39\)

Đẳng thức xảy ra khi a = 6

B3: Áp dụng bđt AM-GM

\(A=\frac{a+b}{4\sqrt{ab}}+\frac{\sqrt{ab}}{a+b}+\frac{3\left(a+b\right)}{4\sqrt{ab}}\ge2\sqrt{\frac{a+b}{4\sqrt{ab}}.\frac{\sqrt{ab}}{a+b}}+\frac{3\left(a+b\right)}{4\left(\frac{a+b}{2}\right)}\)

\(=1+\frac{3\left(a+b\right)}{2\left(a+b\right)}=1+\frac{3}{2}=\frac{5}{2}\)

Dấu "=" xảy ra khi \(a=b>0\)