\(M=a+\dfrac{1}{a}=\dfrac{3a}{4}+\dfrac{a}{4}+\dfrac{1}{a}\)

BBĐT AM-GM

\(=>\dfrac{a}{4}+\dfrac{1}{a}\ge2\sqrt{\dfrac{1}{4}}=1\)

\(=>M=\dfrac{3a}{4}+\dfrac{a}{4}+\dfrac{1}{a}\ge1+\dfrac{3.2}{4}=\dfrac{5}{2}\)

dấu"=" xảy ra<=>\(a=2\)

cánh 2: \(M=a+\dfrac{1}{a}\ge2+\dfrac{1}{2}=\dfrac{5}{2}\) dấu"=" xảy ra tương tự

\(A=2n^2\left(2n-1\right)-3\left(2n-1\right)+2=\left(2n^2-3\right)\left(2n-1\right)+2\)

Do \(\left(2n^2-3\right)\left(2n-1\right)⋮2n-1\)

\(\Rightarrow2⋮2n-1\)

\(\Rightarrow2n-1=Ư\left(2\right)\)

Mà 2n-1 luôn lẻ \(\Rightarrow2n-1=\left\{-1;1\right\}\)

\(\Rightarrow n=\left\{0;1\right\}\)

2.

\(Q=-\left(x^2+4x+4\right)-\left(y^2-2y+1\right)+7\)

\(Q=-\left(x+2\right)^2-\left(y-1\right)^2+7\le7\)

\(Q_{max}=7\) khi \(\left(x;y\right)=\left(-2;1\right)\)

\(cos\left(2-ab\right)-cos\left(a+b\right)=a+b+ab-2\)

\(\Leftrightarrow cos\left(2-ab\right)+2-ab=cos\left(a+b\right)+a+b\)

Xét hàm \(f\left(x\right)=cosx+x\)

\(f'\left(x\right)=-sinx+1\ge0;\forall x\Rightarrow f\left(x\right)\) đồng biến trên R

\(\Rightarrow2-ab=a+b\)

\(\Rightarrow2-a=b\left(a+1\right)\Rightarrow b=\dfrac{2-a}{a+1}=\dfrac{3}{a+1}-1\)

\(\Rightarrow P=a+\dfrac{6}{a+1}-2=a+1+\dfrac{6}{a+1}-3\ge2\sqrt{\dfrac{6\left(a+1\right)}{a+1}}-3=2\sqrt{6}-3\)

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

\(\sqrt{a^2+\dfrac{1}{b+c}}=\dfrac{2}{\sqrt{17}}\sqrt{\left(4+\dfrac{1}{4}\right)\left(a^2+\dfrac{1}{b+c}\right)}\ge\dfrac{2}{\sqrt{17}}\left(2a+\dfrac{1}{2\sqrt{b+c}}\right)\)

\(\Rightarrow A\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{1}{\sqrt{a+b}}+\dfrac{1}{\sqrt{b+c}}+\dfrac{1}{\sqrt{c+a}}\right)\)

\(\Rightarrow A\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}\right)\)

Mặt khác:

\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{3\left(a+b+b+c+c+a\right)}=\sqrt{6\left(a+b+c\right)}\)

\(\Rightarrow A\ge\dfrac{1}{\sqrt{17}}\left(4a+4b+4c+\dfrac{9}{\sqrt{6\left(a+b+c\right)}}\right)\)

\(\Rightarrow A\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}\left(a+b+c\right)+\dfrac{a+b+c}{8}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}+\dfrac{9}{2\sqrt{6\left(a+b+c\right)}}\right)\)

\(\Rightarrow A\ge\dfrac{1}{\sqrt{17}}\left(\dfrac{31}{8}.6+3\sqrt[3]{\dfrac{81\left(a+b+c\right)}{32.6.\left(a+b+c\right)}}\right)=\dfrac{3\sqrt{17}}{2}\)

Dấu "=" xảy ra khi \(a=b=c=2\)

cho mk hỏi kq co pkai la Pmin = \(\sqrt{2}\) khi và chỉ khi a=b pkai k ạ . bài này mk đang cần gấp ạ nên bạn nào đi qua đọc dk thì xin bạn hãy viết cách giải lên hộ mk . để mk có thể  xem  cách làm của mk đúng hay chưa ạ ( mk giải theo BDT cosi ạ )

\(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge2\sqrt{ab}.2\sqrt{\frac{1}{a}.\frac{1}{b}}=4\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}=\frac{4}{2\sqrt{2}}=\sqrt{2}\)

Dấu bằng xảy ra khi \(a=b=\sqrt{2}\)

Câu a : Áp dụng BĐT Cô - si cho các số dương ta có :

\(\left\{{}\begin{matrix}\frac{a^2}{b}+b\ge2\sqrt{\frac{a^2}{b}.b}=2a\\\frac{b^2}{c}+c\ge2\sqrt{\frac{b^2}{c}.c}=2b\\\frac{c^2}{a}+a\ge2\sqrt{\frac{c^2}{a}.a}=2c\end{matrix}\right.\)

Cộng từng vế của BĐT ta thu được :

\(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+a+b+c\ge2a+2b+c\)

\(\Leftrightarrow\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c=1\) ( đpcm )

Dấu \("="\) xảy ra khi \(a=b=c=\frac{1}{3}\)

\(M\ge\frac{\left(1+1+1+1\right)^2}{3\left(a+b+c+d\right)}=\frac{16}{3\left(a+b+c+d\right)}\) ( bdt Cauchy dạng Engel)

Mặt khác, có \(\left(a+b+c+d\right)^2\le4\left(a^2+b^2+c^2+d^2\right)\le16\) ( bdt Bunykovski)

\(\Leftrightarrow a+b+c+d\le4\)

\(\Rightarrow M\ge\frac{16}{3\left(a+b+c+d\right)}\ge\frac{16}{12}=\frac{4}{3}\)

Dấu "=" : x =y =z = 1