Làm nhầm đề \(4ac^2\) mất nửa tiếng mãi không ra, đề cho dễ nhầm lẫn quá.

Ta có:

\(P=a^2b-abc+c\left(2a-b\right)^2\ge a^2b-abc=ab\left(a-c\right)\)

- Nếu \(a>c\Rightarrow P\ge0\)

- Nếu \(a\le c\Rightarrow P\ge ab\left(a-c\right)=-\dfrac{1}{2}.2a.b\left(c-a\right)\)

\(\Rightarrow P\ge-\dfrac{1}{54}\left(2a+b+c-a\right)^3=-4\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(1;2;3\right)\)

Max:

- Nếu \(b>a\):

\(P=a^2b+b^2c+4ca^2-5abc< ab^2+b^2c+ca\left(4a-5b\right)< ab^2+b^2c\)

\(P< b^2\left(a+c\right)=4.\dfrac{b}{2}.\dfrac{b}{2}\left(a+c\right)\le\dfrac{4}{27}\left(\dfrac{b}{2}+\dfrac{b}{2}+a+c\right)^3=32\)

- Nếu \(b\le a\):

\(P=a^2b+b^2c+4ca^2-5abc\le4a^2b+4b^2c+4ca^2-4abc\)

\(P\le4a^2\left(b+c\right)+4bc\left(b-a\right)\le4a^2\left(b+c\right)\)

\(P\le16.\dfrac{a}{2}.\dfrac{a}{2}\left(b+c\right)\le\dfrac{16}{27}\left(\dfrac{a}{2}+\dfrac{a}{2}+b+c\right)^3=128\)

Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(4;0;2\right)\)

P/s: mình sẽ ko làm những bài BĐT nhiều hơn 3 biến hoặc các dạng tổng quát (phí thời gian).

a.

\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)

2.

\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)

Quay lại câu a

\(b,\dfrac{ab}{a+3b+2c}=\left(\dfrac{1}{9}ab\right)\cdot\dfrac{9}{\left(a+c\right)+\left(b+c\right)+2b}\le\left(\dfrac{1}{9}ab\right)\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\cdot\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)

Cmtt: \(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\cdot\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+b}+\dfrac{b}{2}\right);\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\cdot\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)

\(\Rightarrow VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ab+ac}{b+c}+\dfrac{ab+bc}{a+c}+\dfrac{a+b+c}{2}\right)\\ \le\dfrac{1}{9}\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{9}\cdot\dfrac{3}{2}\left(a+b+c\right)=\dfrac{a+b+c}{6}\)

Dấu $"="$ khi $a=b=c$

xét hiệu \(\frac{1}{a}+\frac{1}{b}-\frac{4}{a+b}=\frac{a+b}{ab}-\frac{4}{a+b}\)

\(=\frac{\left(a+b\right)^2}{ab\left(a+b\right)}-\frac{4ab}{ab\left(a+b\right)}\)

\(=\frac{a^2+2ab+b^2-4ab}{ab\left(a+b\right)}\)

\(=\frac{\left(a-b\right)^2}{ab\left(a+b\right)}\)

vì (a-b)²>=0

mà a,b>0 nên ab>0;a+b>0

\(\Rightarrow\frac{\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\frac{1}{a}+\frac{1}{b}-\frac{4}{ab}\ge0\)

hay \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{ab}\left(dpcm\right)\)

c) Áp dụng BĐT cô si cho 2 hai số dương \(a;b\) ta có:

\(a+b\ge2\sqrt{ab}\)

\(\frac{1}{a}+\frac{1}{b}\ge\frac{1}{\sqrt{ab}}\)

\(\Rightarrow\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)

\(\Rightarrow\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)

Dấu "=" xảy ra khi \(\Leftrightarrow a=b\)

\(\frac{a^6}{b^2}+\frac{b^6}{a^2}=\frac{a^8+b^8}{a^2b^2}\ge\frac{\left(a^4+b^4\right)^2}{2a^2b^2}=\frac{\left(a^4+b^4\right)\left(a^4+b^4\right)}{2a^2b^2}\ge\frac{\left(a^4+b^4\right).2a^2b^2}{2a^2b^2}=a^4+b^4\)

Dấu "=" xảy ra khi \(a^2=b^2\)

Ta có:

\(\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)

\(\Leftrightarrow\dfrac{3a\left(a+b\right)+3b\left(a+b\right)-12ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{3a^2+3ab+3ab+3b^2-12ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{3a^2+3b^2-6ab}{ab\left(a+b\right)}\ge0\)

\(\Leftrightarrow\dfrac{3\left(a-b\right)^2}{ab\left(a+b\right)}\ge0\) ( luôn đúng)

Tương tự ta có:

\(\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)

\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)

Cộng vế (1) (2)(3) ta được:

\(\dfrac{3}{a}+\dfrac{3}{b}+\dfrac{2}{b}+\dfrac{2}{c}+\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{12}{a+b}+\dfrac{8}{b+c}+\dfrac{4}{c+a}\)

\(\Leftrightarrow\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\)

1/a+1/b>=2/căn ab

a+b>=2căn ab

=>(1/a+1/b)(a+b)>=4

Từ BĐT trên ,ta có:

\(\dfrac{1}{a}\)+\(\dfrac{1}{b}\) \(\geq\) \(\dfrac{4}{a+b}\)

\(\Leftrightarrow\) \(\dfrac{a+b}{ab}\) \(\geq\) \(\dfrac{4}{a+b}\)

\(\Leftrightarrow\) (a+b)(a+b) \(\geq\) 4ab

\(\Leftrightarrow\) (a+b)² \(\geq\) 4ab

\(\Leftrightarrow\) a² +2ab+b²\(\geq\) 4ab

\(\Leftrightarrow\) a²+2ab+b²-4ab \(\geq\) 0

\(\Leftrightarrow\) a²-2ab+b² \(\geq\) 0

\(\Leftrightarrow\) (a-b)² \(\geq\) 0 (luôn đúng)

Dấu '=' xảy ra khi và chỉ khi a=b

Từ đó ta chứng minh được BĐT : \(\dfrac{1}{a}\) +\(\dfrac{1}{b}\)\(\geq\) \(\dfrac{4}{a+b}\)

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

\(\dfrac{4}{a+b}=\dfrac{4ab}{ab\left(a+b\right)}\) (2)

ta có:

\(\left(a+b\right)^2\ge\left(a-b\right)^2\) và \(\left(a-b\right)^2\ge4ab\)

nên \(\left(a+b\right)^2\ge4ab\) (3)

từ (1), (2) và (3) suy ra \(\dfrac{\left(a+b\right)^2}{ab\left(a+b\right)}\ge\dfrac{4ab}{ab\left(a+b\right)}\) hay \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)(đpcm)

Xét hiệu : VT - VP

= \(\dfrac{1}{a}\) + \(\dfrac{1}{b}\) - \(\dfrac{4}{a+b}\) = \(\dfrac{a+b}{ab}\) - \(\dfrac{4}{a+b}\)

= \(\dfrac{\left(a+b\right)^2}{ab.\left(a+b\right)}\) - \(\dfrac{4ab}{ab.\left(a+b\right)}\)

= \(\dfrac{a^2+2ab+b^2-4ab}{ab.\left(a+b\right)}\) = \(\dfrac{\left(a-b\right)^2}{ab.\left(a+b\right)}\)

Vì a,b > 0 => a.b > 0 ; a + b > 0

=> ab.(a + b) > 0 . Mà ( a - b)² > 0 nên

\(\dfrac{\left(a-b\right)^2}{ab.\left(a+b\right)}\) \(\ge\) 0 hay VT - VP \(\ge\) 0 ( BĐT đúng )

=> \(\dfrac{1}{a}\) + \(\dfrac{1}{b}\) \(\ge\) \(\dfrac{4}{a+b}\)

1) Sửa lại:Cho x,y,z dương nhé!

\(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=x\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+y\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+z\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

\(=1+\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+1+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}+1=\left(1+1+1\right)+\left(\frac{x}{y}+\frac{x}{z}+\frac{y}{x}+\frac{y}{z}+\frac{z}{x}+\frac{z}{y}\right)\)

\(=3+\left(\frac{x}{y}+\frac{y}{x}\right)+\left(\frac{y}{z}+\frac{z}{y}\right)+\left(\frac{z}{x}+\frac{x}{z}\right)\)

Vì x,y,z là các số dương ,ta áp dụng bất đẳng thức Cô-Si:

\(\frac{x}{y}+\frac{y}{x}\ge2\sqrt{\frac{x}{y}.\frac{y}{x}}=2\)

\(\frac{y}{z}+\frac{z}{y}\ge2\sqrt{\frac{y}{z}.\frac{z}{y}}=2\)

\(\frac{z}{x}+\frac{x}{z}\ge2\sqrt{\frac{z}{x}.\frac{x}{z}}=2\)

Do đó \(\left(x+y+z\right)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge3+2+2+2=9\)

Dấu "=" xảy ra <=> \(x=y=z\)

câu 2) mk chịu

câu 2 đề sai . sửa số 3 thành số 2 . neu sua thanh co 2 thi co the ap dung bdt cosi hoac trebusep