Áp dụng bất đẳng thức Svacxo ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+2b}\)

Tương tự : \(\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{9}{b+2c};\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{9}{c+2a}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{a+2b}+\dfrac{3}{b+2c}+\dfrac{3}{c+2a}\)

Dấu = xảy ra khi a=b=c

\(=>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+2b}\)(BĐT Cauchy Schawarz)(1)

tương tự \(=>\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{c}\ge\dfrac{9}{b+2c}\left(2\right)\)

\(=>\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{a}\ge\dfrac{9}{c+2a}\left(3\right)\)

(1)(2)(3)

\(=>3\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)

\(=>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\left(dpcm\right)\)

Chứng minh biểu thức \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{b}\ge\dfrac{9}{a+2b}\\ \Leftrightarrow\dfrac{a+2b}{a}+\dfrac{2\left(a+2b\right)}{b}\ge9\\ \Leftrightarrow\dfrac{2b}{a}+\dfrac{2a}{b}\ge4\\ \Leftrightarrow\dfrac{a}{b}+\dfrac{b}{a}\ge2\left(cosi\right)\)

Áp dụng BĐT \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{9}{x+y+z}\) ta được

\(\dfrac{1}{2a}+\dfrac{1}{2b}+\dfrac{1}{2b}\ge\dfrac{9}{2\left(a+2b\right)}\)

\(\dfrac{1}{2b}+\dfrac{1}{2c}+\dfrac{1}{2c}\ge\dfrac{9}{2\left(b+2c\right)}\)

\(\dfrac{1}{2c}+\dfrac{1}{2a}+\dfrac{1}{2a}\ge\dfrac{9}{2\left(c+2a\right)}\)

Cộng các BĐT theo vế

\(\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{9}{2}\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge3\left(\dfrac{1}{a+2b}+\dfrac{1}{b+2c}+\dfrac{1}{c+2a}\right)\)

Dấu " = " xảy ra khi a = b = c ( a,b,c > 0 )

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+c}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c\left(b+c\right)}{8a^3\left(b+c\right)b^2c}}=\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{c+a}{4ca}+\dfrac{1}{2c}\ge3\sqrt[3]{\dfrac{c^2a\left(c+a\right)}{8b^3\left(c+a\right)c^2a}}=\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{a+b}{4ab}+\dfrac{1}{2a}\ge3\sqrt[3]{\dfrac{a^2b\left(a+b\right)}{8c^3\left(a+b\right)a^2b}}=\dfrac{3}{2c}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{1}{4b}+\dfrac{1}{2b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{1}{4c}+\dfrac{1}{2c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{1}{4a}+\dfrac{1}{2a}\ge\dfrac{3}{2c}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{1}{4c}+\dfrac{3}{4b}\ge\dfrac{3}{2a}\\\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{1}{4a}+\dfrac{3}{4c}\ge\dfrac{3}{2b}\\\dfrac{a^2b}{c^3\left(a+b\right)}+\dfrac{1}{4b}+\dfrac{3}{4a}\ge\dfrac{3}{2c}\end{matrix}\right.\)

\(\Rightarrow VT+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{3}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Rightarrow VT+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{3}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Rightarrow VT\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(\Leftrightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ( đpcm )

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 \(a=b=1\)

Đề bài sai với \(a=b=c=2\)

đề đúng nhớ áp dụng AM-GM

AD bđt AM-GM cho 3 số

\(\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{b+C}{4bc}+\dfrac{1}{2b}\ge3\sqrt[3]{\dfrac{b^2c}{a^3\left(b+c\right)}.\dfrac{\left(b+c\right)}{4bc}.\dfrac{1}{2b}}=\dfrac{3}{2a}\)

\(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}\ge\dfrac{3}{2a}-\dfrac{3}{4b}-\dfrac{1}{4c}\)

thiết lập bđt tương tự r cộng lại \(\Rightarrow\dfrac{b^2c}{a^3\left(b+c\right)}+\dfrac{c^2a}{b^3\left(c+a\right)}+\dfrac{a^2b}{c^3\left(a+b\right)}\ge\left(\dfrac{3}{2}-\dfrac{3}{4}-\dfrac{1}{4}\right)\left(a+b+c\right)=\dfrac{1}{2}\left(a+b+c\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}a^2b+\dfrac{1}{b}\ge2\sqrt{\dfrac{a^2b}{b}}=2a\\b^2c+\dfrac{1}{c}\ge2\sqrt{\dfrac{b^2c}{c}}=2b\\c^2a+\dfrac{1}{a}\ge2\sqrt{\dfrac{c^2a}{a}}=2c\end{matrix}\right.\)

\(\Rightarrow a^2b+b^2c+c^2a+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge2\left(a+b+c\right)\)

\(\Rightarrow\dfrac{1}{2}\left(a^2b+b^2c+c^2a+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge a+b+c\) ( đpcm )

Dấu " = " xảy ra khi \(a=b=c=1\)