\(S=a^2+\dfrac{1}{a^2}\)

\(S=\dfrac{1}{16}a^2+\dfrac{1}{a^2}+\dfrac{15}{16}a^2\)

\(S\ge2\sqrt{\dfrac{1}{16}a^2\cdot\dfrac{1}{a^2}}+\dfrac{15}{16}\cdot2^2\)

\(S\ge2\cdot\dfrac{1}{4}+\dfrac{15}{4}\)

\(S\ge\dfrac{17}{4}\)

Vậy \(MINS=\dfrac{17}{4}\Leftrightarrow a=2\)

\(P=16\left(a-\frac{1}{2}\right)^2+2\left(b-1\right)^2+\left(\frac{3}{a}+12a\right)+\left(\frac{2}{b}+2b\right)+2\left(2a+b\right)-6\ge14\)

"=" \(\Leftrightarrow\)\(a=\frac{1}{2};b=1\)

Sorry.Because I am thirty years old 

vc 30 tuổi mà ko bik làm

Tuổi ảo mà

Ta đặt:

     \(\left\{{}\begin{matrix}x=a-1\\y=b-2\\z=c-3\end{matrix}\right.\)

        \(\Rightarrow x+y+z=3\) và  \(x,y,z\ge0\) (*)

Biểu thứ P trở thành:

     \(P=\sqrt{x}+\sqrt{y}+\sqrt{z}\)

Từ (*) dễ thấy:

     \(\left\{{}\begin{matrix}0\le x\le3\\0\le y\le3\\0\le z\le3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\le x\le\sqrt{3x}\\0\le y\le\sqrt{3y}\\0\le z\le\sqrt{3z}\end{matrix}\right.\)

Do đó:

     \(P\ge\dfrac{x+y+z}{\sqrt{3}}=\sqrt{3}\)

Dầu "=" xảy ra khi \(\left(a;b;c\right)=\left(3;0;0\right)=\left(0;3;0\right)=\left(0;0;3\right)\)

\(S=\frac{a}{8}+\frac{a}{8}+\frac{1}{a^2}+\frac{3a}{4}\ge3\sqrt[3]{\frac{a.a.1}{8.8.a^2}}+\frac{3}{4}.2=\frac{9}{4}\)

Min S = 9/4 khi a =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 \(Min_S=\frac{27}{4}\)

tôi cho sai đó

 Tô Quang Huy :Làm được thì làm, ko làm được đừng có trả lời linh tinh, đinh spam à == ko làm được thì đừng nói vớ vẩn, là mất thgian của người khác ạ:(

Bài làm:

a) \(A=x+\frac{1}{2}\ge2+\frac{1}{2}=\frac{5}{2}\)

Dấu "=" xảy ra khi x = 2

Vậy \(Min\left(A\right)=\frac{5}{2}\)khi x = 2

b) \(B=x+\frac{3}{x^2}=\left(\frac{3}{x^2}+\frac{3x}{8}+\frac{3x}{8}\right)+\frac{x}{4}\ge3\sqrt[3]{\frac{3}{x^2}.\frac{3x}{8}.\frac{3x}{8}}+\frac{2}{4}\)

\(=3.\frac{3}{4}+\frac{1}{2}=\frac{11}{4}\)

Dấu "=" xảy ra khi x = 2

Vậy \(Min\left(B\right)=\frac{11}{4}\)khi x = 2

c) \(C=2x+\frac{3}{x}+\frac{4}{x^2}=\left(\frac{4}{x^2}+\frac{x}{2}+\frac{x}{2}\right)+\left(\frac{3x}{4}+\frac{3}{x}\right)+\frac{x}{4}\)

\(\ge3.\sqrt[3]{\frac{4}{x^2}.\frac{x}{2}.\frac{x}{2}}+2.\sqrt{\frac{3x}{4}.\frac{3}{x}}+\frac{2}{4}\)

\(=3.1+2.\frac{3}{2}+\frac{1}{2}=\frac{13}{2}\)

Dấu "=" xảy ra khi x = 2

Vậy \(Min\left(C\right)=\frac{13}{2}\)khi x = 2

Học tốt!!!!