\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

=> bc+ac+ab=0

ta có

\(bc+ac=-ab\)

<=> \(\left(bc+ac\right)^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2+2abc^2=a^2b^2\)

<=> \(b^2c^2+a^2c^2-a^2b^2=-2abc^2\)

tương tự

\(a^2b^2+b^2c^2-c^2a^2=-2ab^2c\)

\(c^2a^2+a^2b^2-b^2c^2=-2a^2bc\)

thay vào E ta đc

\(E=\dfrac{-a^2b^2c^2}{2ab^2c}-\dfrac{a^2b^2c^2}{2abc^2}-\dfrac{a^2b^2c^2}{2a^2bc}\)

=\(-\dfrac{ac}{2}-\dfrac{ab}{2}-\dfrac{bc}{2}=\dfrac{-\left(ac+ab+bc\right)}{2}=0\) (vì ac+bc+ab=0 cmt)

Hình như sai đề :

Ta có : \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

\(\Leftrightarrow\dfrac{bc}{abc}+\dfrac{ac}{abc}+\dfrac{ab}{abc}=0\)

\(\Leftrightarrow\dfrac{ab+ac+bc}{abc}=0\)

\(\Leftrightarrow ab+ac+bc=0\) ( do \(a;b;c\ne0\) ) ( 1 )

Từ ( 1 ) \(\Rightarrow ab+bc=-ac\)

\(\Rightarrow\left(ab+bc\right)^2=\left[-\left(ac\right)\right]^2\)

\(\Rightarrow a^2b^2+b^2c^2+2ab^2c=a^2c^2\) ( * )

CMTT , ta được : \(\left\{{}\begin{matrix}b^2c^2+c^2a^2+2bc^2a=a^2b^2\\c^2a^2+a^2b^2+2a^2cb=b^2c^2\end{matrix}\right.\) ( *' )

Thay ( * ) và ( * ') vào E , ta được :

\(E=\dfrac{a^2b^2c^2}{a^2b^2+b^2c^2-\left(a^2b^2+b^2c^2+2b^2ac\right)}+\dfrac{a^2b^2c^2}{b^2c^2+c^2a^2-\left(b^2c^2+c^2a^2+2bc^2a\right)}\)

\(+\dfrac{a^2b^2c^2}{c^2a^2+a^2b^2-\left(c^2a^2+a^2b^2+2a^2cb\right)}\)

\(=\dfrac{a^2b^2c^2}{-2b^2ac}+\dfrac{a^2b^2c^2}{-2c^2ab}+\dfrac{a^2b^2c^2}{-2a^2cb}\)

\(=\dfrac{-ac}{2}+\dfrac{-ab}{2}+\dfrac{-bc}{2}\)

\(=\dfrac{-\left(ac+ab+bc\right)}{2}\)

\(=\dfrac{0}{2}=0\)

Vậy \(E=0\)

\(a+b+c=\sqrt{6063}\Leftrightarrow\dfrac{a}{\sqrt{2021}}+\dfrac{b}{\sqrt{2021}}+\dfrac{c}{\sqrt{2021}}=\sqrt{3}\)

Đặt \(\left(\dfrac{a}{\sqrt{2021}};\dfrac{b}{\sqrt{2021}};\dfrac{c}{\sqrt{2021}}\right)=\left(x;y;z\right)\Rightarrow x+y+z=\sqrt{3}\)

\(P=\dfrac{2x}{\sqrt{2x^2+1}}+\dfrac{2y}{\sqrt{2y^2+1}}+\dfrac{2z}{\sqrt{2z^2+1}}\)

Ta có đánh giá:

\(\dfrac{x}{\sqrt{2x^2+1}}\le\dfrac{3\sqrt{15}x+2\sqrt{5}}{25}\)

Thật vậy, BĐT tương đương:

\(\left(\sqrt{3}x-1\right)^2\left(9x^2+10\sqrt{3}x+2\right)\ge0\) (luôn đúng)

Tương tự và cộng lại:

\(P\le\dfrac{6\sqrt{15}\left(x+y+z\right)+12\sqrt{5}}{25}=\dfrac{6\sqrt{5}}{5}\)

Áp dụng BĐt cô-si, ta có \(\frac{2\left(a+b\right)^2}{2a+3b}\ge\frac{8ab}{2a+3b}=\frac{8}{\frac{2}{b}+\frac{3}{a}}\)

                                      \(\frac{\left(b+2c\right)^2}{2b+c}\ge\frac{8bc}{2b+c}=\frac{8}{\frac{2}{c}+\frac{1}{b}}\)

                                        \(\frac{\left(2c+a\right)^2}{c+2a}\ge\frac{8ac}{c+2a}\ge\frac{8}{\frac{1}{a}+\frac{2}{c}}\)

Cộng 3 cái vào, ta có 

A\(\ge8\left(\frac{1}{\frac{2}{b}+\frac{3}{a}}+\frac{1}{\frac{1}{b}+\frac{2}{c}}+\frac{1}{\frac{1}{a}+\frac{2}{c}}\right)\ge8\left(\frac{9}{\frac{3}{b}+\frac{4}{c}+\frac{4}{a}}\right)=8.\frac{9}{3}=24\)

Vậy A min = 24 

Neetkun ^^

bạn tìm ra dấu= xảy ra khi nào

rất tiếc sai rồi :)) 

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

Tương tự: \(\dfrac{\sqrt{c^2+2b^2}}{bc}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{b}+\dfrac{2}{c}\right)\) ; \(\dfrac{\sqrt{a^2+2c^2}}{ac}\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{1}{c}+\dfrac{2}{a}\right)\)

Cộng vế với vế:

\(VT\ge\dfrac{1}{\sqrt{3}}\left(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{3}{c}\right)=\sqrt{3}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=1980\sqrt{3}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{3}{1980}\)

Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) 

Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)

CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)

\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)

Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu = xảy ra khi a=b=c=3

Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)

\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)

\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)

Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)

 \(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)

\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)

Vậy...

2,

\(ab\le\dfrac{1}{4}\left(a+b\right)^2=1\Rightarrow0\le ab\le1\)

\(E=9a^2b^2+6\left(a^3+b^3\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+5ab\left(a+b\right)+24ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(ab=x\Rightarrow0\le x\le1\)

\(E=9x^2-2x+48=\left(x-1\right)\left(9x+7\right)+55\le55\)

\(E_{max}=55\) khi \(x=1\) hay \(a=b=1\)