Đề bài sai

Phản ví dụ: \(a=\dfrac{1}{2};b=2;c=4\) vì VT<VP

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Khi đó, ta có : \(\frac{3bk+2b}{2bk+3b}=\frac{\left(3k+2\right)b}{\left(2k+3\right)b}=\frac{3k+2}{2k+3}\)(1)

       \(\frac{3dk+2d}{2dk+3d}=\frac{\left(3k+2\right).d}{\left(2k+3\right).d}=\frac{3k+2}{2k+3}\)(2)

Từ (1) và (2), suy ra :  \(\frac{3a+2b}{2a+3b}=\frac{3c+2d}{2c+3d}\)

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 \(a=b=1\)

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

Tương tự và cộng lại;

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{bc}{1+b+bc}+\dfrac{ca}{1+c+ca}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{abc}{a+ab+abc}+\dfrac{ab.ca}{ab+abc+ab.ca}\right)\)

\(P\le\dfrac{1}{2}\left(\dfrac{ab}{1+a+ab}+\dfrac{1}{a+ab+1}+\dfrac{a}{ab+1+a}\right)=\dfrac{1}{2}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(\left\{{}\begin{matrix}a;b;c\ge0\\a+b+c=1\end{matrix}\right.\) \(\Rightarrow0\le a;b;c\le1\)

\(\Rightarrow a\left(a-1\right)\le0\Rightarrow a^2\le a\)

\(\Rightarrow\sqrt{2a^2+3a+4}=\sqrt{a^2+a^2+3a+4}\le\sqrt{a^2+a+3a+4}=a+2\)

Tương tự và cộng lại:

\(\Rightarrow M\le a+2+b+2+c+2=7\)

\(M_{max}=7\) khi \(\left(a;b;c\right)=\left(0;0;1\right)\) và các hoán vị

Sửa đề: CMR: \(\frac{a^2}{2a+3b}+\frac{b^2}{2b+3c}+\frac{c^2}{2c+3a}\ge\frac{1}{5}\left(a+b+c\right)\)

Chứng minh BĐT phụ:

  \(\frac{x^2}{m}+\frac{y^2}{n}\ge\frac{\left(x+y\right)^2}{m+n}\)\(\forall m;n>0\)Tự chứng minh

Áp dụng bđt trên, ta có

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

Vậy..........

Bất đẳng thức cần chứng minh tương đương với:

\(a^3b^2-a^2b^3+b^3c^2-c^3b^2+c^3a^2-c^2a^3\ge0\)

\(\Leftrightarrow a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-a\right)\ge0\)

\(\Leftrightarrow a^2b^2\left(a-b\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-b+b-a\right)\ge0\)

\(\Leftrightarrow a^2b^2\left(a-b\right)+c^2a^2\left(b-a\right)+b^2c^2\left(b-c\right)+c^2a^2\left(c-b\right)\ge0\)

\(\Leftrightarrow\left(a^2b^2-c^2a^2\right)\left(a-b\right)+\left(b^2c^2-c^2a^2\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow a^2\left(b^2-c^2\right)\left(a-b\right)+c^2\left(b^2-a^2\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow\left[a^2\left(b+c\right)-c^2\left(a+b\right)\right]\left(a-b\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow\left(a^2b+a^2c-c^2a-c^2b\right)\left(a-b\right)\left(b-c\right)\ge0\)

\(\Leftrightarrow\left[a\left(ab-c^2\right)+c\left(a^2-bc\right)\right]\left(a-b\right)\left(b-c\right)\ge0\) luôn đúng do \(a\ge b\ge c\ge0\)

cảm ơn bạn nhá, bạn trả lời giúp mình mấy câu hỏi về BĐT còn lại của mik đc ko? cảm ơn bn nhiều!