Cách làm dài bạn thông cảm mình  nghĩ được có zậy thui ak :/

Ta có a, b là các số thực dương 

Từ \(a+3b=ab\Leftrightarrow\frac{1}{b}+\frac{3}{a}=1\ge2\sqrt{\frac{3}{ab}}.\)(bất đẳng thức Cauchy cho 2 số không âm)

\(\Leftrightarrow\frac{12}{ab}\le1\Leftrightarrow ab\ge12\)\(\Leftrightarrow84ab-72ab\ge144\Leftrightarrow84ab\ge72\left(ab+2\right)\)

\(\Leftrightarrow\frac{12ab}{ab+2}\ge\frac{72}{7}\left(1\right)\)

Ta có \(P=\frac{a^2}{1+3b}+\frac{9b^2}{1+a}\ge2\sqrt{\frac{a^2}{1+3b}\frac{9b^2}{1+a}}=\frac{6ab}{\sqrt{\left(1+a\right)\left(1+3b\right)}}\)(Bất đẳng thức Cauchy)

                                                      \(\ge\frac{6ab}{\frac{1+a+1+3b}{2}}=\frac{12ab}{a+3b+2}=\frac{12ab}{ab+2}\)(Bất đẳng thức Cauchy ngược dấu )

Kết hợp với (1) ta được :

\(P\ge\frac{12ab}{ab+2}\ge\frac{72}{7}.\)

Vậy giá trị nhỏ nhất của \(P=\frac{72}{7}\Leftrightarrow\hept{\begin{cases}a=3b\\a+3b=ab\end{cases}\Leftrightarrow\hept{\begin{cases}a=6\\b=2\end{cases}.}}\)

Ta có: P= \(2a+3b+\dfrac{1}{a}+\dfrac{4}{b}\) = \(\text{​​}\text{​​}(\dfrac{1}{a}+a)+\left(\dfrac{4}{b}+b\right)+\left(a+2b\right)\)

Ta thấy: \(\text{​​}\text{​​}(\dfrac{1}{a}+a)\ge2\sqrt{\dfrac{1}{a}\cdot a}=2\)

             \(\text{​​}\text{​​}\left(\dfrac{4}{b}+b\right)\ge2\sqrt{\dfrac{4}{b}\cdot b}=4\)

Do đó: P \(\ge2+4+8=14\)

Vậy: P(min)=14  khi:  \(\left\{{}\begin{matrix}\dfrac{1}{a}=a\\\dfrac{4}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=2\end{matrix}\right..\)

sorry, nhầm đề

làm lại:

Ta có: P= \(2a+3b+\dfrac{4}{a}+\dfrac{9}{b}\) = \(\text{​​}\text{​​}(\dfrac{4}{a}+a)+\left(\dfrac{9}{b}+b\right)+\left(a+2b\right)\)

Ta thấy: \(\text{​​}\text{​​}(\dfrac{4}{a}+a)\ge2\sqrt{\dfrac{4}{a}\cdot a}=4\)

             \(\text{​​}\text{​​}\left(\dfrac{9}{b}+b\right)\ge2\sqrt{\dfrac{9}{b}\cdot b}=6\)

Do đó: P \(\ge4+6+8=18\)

Vậy: P(min)=18  khi:  \(\left\{{}\begin{matrix}\dfrac{4}{a}=a\\\dfrac{9}{b}=b\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=3\end{matrix}\right.\cdot\)

Bài làm

\(P=2a+3b+\frac{4}{a}+\frac{9}{b}=a+a+2b+b+\frac{4}{a}+\frac{9}{b}\)

\(=\left(a+2b\right)+\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)\)

\(\ge8+2\sqrt{a\times\frac{4}{a}}+2\sqrt{b\times\frac{9}{b}}\)( Cauchy )

\(=8+4+6=18\)

Đẳng thức xảy ra khi a = 2 ; b = 3

=> MinP = 18 <=> a = 2 ; b = 3

\(P=2a+3b+\frac{4}{a}+\frac{9}{b}\)

\(\Leftrightarrow P=\left(a+\frac{4}{a}\right)+\left(b+\frac{9}{b}\right)+a+2b\)

Áp dụng BĐT AM-GM ta có:

\(P\ge2.\sqrt{a.\frac{4}{a}}+2.\sqrt{b.\frac{9}{b}}+a+2b=2.2+2.3+a+2b\ge4+6+8=18\)

Dấu " = " xảy ra \(\Leftrightarrow\hept{\begin{cases}a=\frac{4}{a}\\b=\frac{9}{b}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)

Vậy \(P_{min}=18\)\(\Leftrightarrow\hept{\begin{cases}a=2\\b=3\end{cases}}\)

\(P=a+2b+a+\frac{4}{a}+b+\frac{9}{b}\)

Áp dụng cô si ta có:

\(a+\frac{4}{a}\ge4\)

\(b+\frac{9}{b}\ge6\)

\(\Rightarrow P\ge4+6+8\Rightarrow P\ge18\)

dấu "=" xảy ra khi a=2,b=3

\(P=2a+3b+\frac{1}{a}+\frac{4}{b}=a+2b+\left(a+\frac{1}{a}\right)+\left(b+\frac{4}{b}\right)\)

   \(\ge5+2\sqrt{a.\frac{1}{a}}+2\sqrt{b.\frac{4}{b}}=5+2+4=11\)

Dấu "=" xảy ra <=>  \(a=1;\)\(b=2\)

Vậy MIN P = 11  Khi a = 1;   b = 2

Bài này là BĐT cosi

\(P=2a+3b+\frac{1}{a}+\frac{4}{b}\)

\(P=a+2b+\left(a+\frac{1}{a}\right)+\left(b+\frac{4}{b}\right)\)

\(P\ge5+2\sqrt{a.\frac{1}{a}}+2\sqrt{b.\frac{4}{b}}=5+2+4=11\)

Dấu "=" xảy ra khi a = 1/a <=> a = 1 ; b = 4/b <=> b = 2

Tại sao lại >= 5 ± √a.1/a vậy

Ta có

\(2a^2\ge8a-8\)(\(2\left(a-2\right)^2\ge0\))

\(7a+\frac{28}{a}\ge28\)

\(b+\frac{1}{b}\ge2\)

\(b^2\ge2b-1\)

Khi đó

\(P\ge a+b+21\ge24\)

Vậy MinP=24 khi a=2, b=1

CÁCH KHÁC:

\(P=\left(2a^2-8a+8\right)+\left(b^2-2b+1\right)+\left(7a+\frac{28}{a}\right)+\left(b+\frac{1}{b}\right)+\left(a+b\right)-9\)

     \(=2\left(a-2\right)^2+\left(b-1\right)^2+\left(7a+\frac{28}{a}\right)+\left(b+\frac{1}{b}\right)+\left(a+b\right)-9\)

       \(\ge2\sqrt{7a.\frac{28}{a}}+2\sqrt{b.\frac{1}{b}}+3-9=24\)

P=(2a2−8a+8)+(b2−2b+1)+(7a+a28​)+(b+b1​)+(a+b)−9

     =2\left(a-2\right)^2+\left(b-1\right)^2+\left(7a+\frac{28}{a}\right)+\left(b+\frac{1}{b}\right)+\left(a+b\right)-9=2(a−2)2+(b−1)2+(7a+a28​)+(b+b1​)+(a+b)−9

       \ge2\sqrt{7a.\frac{28}{a}}+2\sqrt{b.\frac{1}{b}}+3-9=24≥27a.a28​​+2b.b1​​+3−9=24

Ta có: \(\frac{1+3a}{1+b^2}=\left(1+3a\right).\frac{1}{1+b^2}=\left(1+3a\right)\left(1-\frac{b^2}{1+b^2}\right)\)

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

\(=3a+1-\frac{b}{2}-\frac{3ab}{2}\)(1)

Tương tự ta có: \(\frac{1+3b}{1+c^2}=3b+1-\frac{c}{2}-\frac{3bc}{2}\)(2); \(\frac{1+3c}{1+a^2}=3c+1-\frac{a}{2}-\frac{3ca}{2}\)(3)

Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{1+3a}{1+b^2}+\frac{1+3b}{1+c^2}+\frac{1+3c}{1+a^2}\)\(\ge3\left(a+b+c\right)-\frac{a+b+c}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)

\(=\frac{5\left(a+b+c\right)}{2}-\frac{3\left(ab+bc+ca\right)}{2}+3\)

\(\ge\frac{5.\sqrt{3\left(ab+bc+ca\right)}}{2}-\frac{3.3}{2}+3=\frac{15}{2}-\frac{9}{2}+3=6\)

Đẳng thức xảy ra khi a = b = c = 1