\(P=\dfrac{4ab}{a+2b}+\dfrac{9ca}{a+4c}+\dfrac{4bc}{b+c}\)

\(P=\dfrac{4abc}{ac+2bc}+\dfrac{9abc}{ab+4bc}+\dfrac{4abc}{ab+ac}\)

\(P=abc\left(\dfrac{4}{ac+2bc}+\dfrac{9}{ab+4bc}+\dfrac{4}{ab+ac}\right)\)

\(P\ge abc.\dfrac{\left(2+3+2\right)^2}{ac+2bc+ab+4bc+ab+ac}\)

\(P\ge abc.\dfrac{49}{2ab+6bc+2ca}\)

\(P\ge abc.\dfrac{49}{7abc}\) (vì \(2ab+6bc+2ca=7abc\))

\(P\ge7\)

Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2}{ac+2bc}=\dfrac{3}{ab+4bc}=\dfrac{2}{ab+ac}\\2ab+6bc+2ca=7abc\end{matrix}\right.\)

\(\dfrac{2}{ac+2bc}=\dfrac{2}{ab+ac}\) \(\Leftrightarrow2b=a\)

Có \(\dfrac{3}{ab+4bc}=\dfrac{2}{ab+ac}\) 

\(\Leftrightarrow\dfrac{3}{2b^2+4bc}=\dfrac{2}{2b^2+2bc}\) 

\(\Leftrightarrow3b^2+3bc=2b^2+4bc\)

\(\Leftrightarrow b^2=bc\Leftrightarrow b=c\)

\(\Rightarrow a=2b=2c\)

Lại có \(2ab+6bc+2ca=7abc\) \(\Rightarrow4b^2+6b^2+4b^2=14b^3\)

\(\Leftrightarrow b=1\)

\(\Leftrightarrow\left(a,b,c\right)=\left(2,1,1\right)\)

Vậy \(min_P=7\)
 

Từ giả thiết:

\(a^2=2\left(b^2+c^2\right)\ge\left(b+c\right)^2\Rightarrow\left(\dfrac{a}{b+c}\right)^2\ge1\Rightarrow\dfrac{a}{b+c}\ge1\)

\(P=\dfrac{a}{b+c}+\dfrac{b^2}{bc+ab}+\dfrac{c^2}{ac+bc}\ge\dfrac{a}{b+c}+\dfrac{\left(b+c\right)^2}{a\left(b+c\right)+2bc}\ge\dfrac{a}{b+c}+\dfrac{\left(b+c\right)^2}{a\left(b+c\right)+\dfrac{1}{2}\left(b+c\right)^2}\)

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

Đặt \(\dfrac{a}{b+c}=x\ge1\)

\(\Rightarrow P\ge x+\dfrac{1}{x+\dfrac{1}{2}}=\dfrac{4}{9}\left(x+\dfrac{1}{2}\right)+\dfrac{1}{x+\dfrac{1}{2}}+\dfrac{5}{9}x-\dfrac{2}{9}\)

\(P\ge2\sqrt{\dfrac{4}{9}\left(x+\dfrac{1}{2}\right).\dfrac{1}{\left(x+\dfrac{1}{2}\right)}}+\dfrac{5}{9}.1-\dfrac{2}{9}=\dfrac{5}{3}\)

\(P_{min}=\dfrac{5}{3}\) khi \(x=1\) hay \(a=2b=2c\)

Áp dụng bđt AM - GM:

\(P=3a+3b-1+\left[\left(a+1\right)+b+\dfrac{c^3}{b\left(a+1\right)}\right]\ge3a+3b-1+3c=3.5-1=14\).

Đẳng thức xảy ra khi a = 1; b = 2; c = 2.

Vậy Min P = 14 khi a = 1; b = 2; c = 2.

Chọn D

Chọn đáp án D

Ta có

Suy ra

Từ giả thiết ta có f ' x + f ' ' x = 10 e x  

Để phương trình  f ' x + f ' ' x = 10 e x có nghiệm

⇔  Phương trình (*) có nghiệm

* Nếu b = 0 thì S = a 2 ≥ 10  

* Nếu b ≠ 0 thì S = a 2 - 2 a b + 3 b 2 ≥ 10 . a b 2 - 2 . a b + 3 a b 2 + 1 .

Đặt t = a b t ∈ R , suy ra S ≥ 10 . t 2 - 2 t + 3 t 2 + 1 .

Xét hàm số f t = t 2 - 2 t + 3 t 2 + 1  trên R.

Ta có

Bảng biến thiên:

Quan sát bảng biến thiên ta thấy  f t ≥ 2 - 2

\(abc\ge\left(a+b-c\right)\left(b+c-a\right)\left(c+a-b\right)\)

\(\Leftrightarrow abc\ge\left(3-2a\right)\left(3-2b\right)\left(3-2c\right)\)

\(\Leftrightarrow9abc\ge12\left(ab+bc+ca\right)-27\)

\(\Rightarrow abc\ge\dfrac{4}{3}\left(ab+bc+ca\right)-3\)

\(P\ge\dfrac{9}{a\left(b^2+bc+c^2\right)+b\left(c^2+ca+a^2\right)+c\left(a^2+ab+b^2\right)}+\dfrac{abc}{ab+bc+ca}=\dfrac{9}{\left(ab+bc+ca\right)\left(a+b+c\right)}+\dfrac{abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3}{ab+bc+ca}+\dfrac{abc}{ab+bc+ca}=\dfrac{3+abc}{ab+bc+ca}\)

\(\Rightarrow P\ge\dfrac{3+\dfrac{4}{3}\left(ab+bc+ca\right)-3}{ab+bc+ca}=\dfrac{4}{3}\)

Dấu "=" xảy ra khi \(a=b=c=1\)

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