Ta có: 

\(4\le\left(\sqrt{a}+1\right)\left(\sqrt{b}+1\right)=\sqrt{ab}+\sqrt{a}+\sqrt{b}+1\le\dfrac{a+b}{2}+\dfrac{a+1}{2}+\dfrac{b+1}{2}+1\)

\(=a+b+2\)

\(\Leftrightarrow a+b\ge2\)

\(\dfrac{a^2}{b}+\dfrac{b^2}{a}\ge\dfrac{\left(a+b\right)^2}{a+b}=a+b\ge2\)

Dấu \(=\) xảy ra khi \(a=b=1\).

áp dụng BĐT Cô si :

+ cho cặp số a,b ta được \(a+b\ge2\sqrt{ab}\left(1\right)\)

+ cho cặp số \(\dfrac{1}{a}+\dfrac{1}{b}\) ta được \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{2}{\sqrt{ab}}\left(2\right)\)

Nhân hai vế với \(\left(1\right),\left(2\right)\) ta được :\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge2\sqrt{ab}.\dfrac{2}{\sqrt{ab}}=4\) (đpcm)

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

=> (a+b).\(\left(\dfrac{1}{b}+\dfrac{1}{b}\right)\ge\left(a+b\right).\dfrac{4}{a+b}=4\left(dpcm\right)\)

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

=>\(\left(a+b+c\right).\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(a+b+c\right).\dfrac{9}{a+b+c}=9\left(dpcm\right)\)

oaa cha cha :V mới đọc BĐT kiểu dạng này xong :P

Mình sẽ giải theo hai cách nhé :P

C1 : Áp dụng BĐT Cauchy - Schwarz dạng engel :

\(\dfrac{a^2_1+a^2_2+...+a^2_n}{b_1+b_2+...+b_n}\ge\dfrac{\left(a_1+a_2+...+a_n\right)^2}{b_1+b_2+...+b_n}\) Ta có :

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{\left(1+1\right)^2}{a+b}=\dfrac{4}{ab}\left(ĐPCM\right)\)

Đẳng thức xảy ra khi \(\dfrac{1}{a}=\dfrac{1}{b}\)

C2 : Áp dụng BĐT Cauchy dạng \(a+b\ge2ab\) ta có :

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

\(=1+1+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\ge2+2\sqrt{\dfrac{a}{b}\cdot\dfrac{b}{a}}=2+2\sqrt{1}=4\left(ĐPCM\right)\)

Đẳng thức xảy ra khi a = b.

vì a,b>0, áp dụng bđt cô si ta có

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

\(\dfrac{1}{a}+\dfrac{1}{b}\ge2\sqrt{\dfrac{1}{ab}}\)

nhân với nhau ta có

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

Lời giải:

Bổ sung điều kiện $a,b$ là các số dương. Áp dụng BĐT Cô-si ta có:

$a+b\geq 2\sqrt{ab}$

$\frac{1}{a}+\frac{1}{b}\geq 2\sqrt{\frac{1}{ab}}$

$\Rightarrow (a+b)(\frac{1}{a}+\frac{1}{b})\geq 2\sqrt{ab}.2\sqrt{\frac{1}{ab}}=4$

Ta có đpcm 

Dấu "=" xảy ra khi $a=b$

\(\dfrac{a^3}{\left(a+2b\right)\left(b+2c\right)}+\dfrac{a+2b}{27}+\dfrac{b+2c}{27}\ge3\sqrt[3]{\dfrac{a^3\left(a+2b\right)\left(b+2c\right)}{27^2.\left(a+2b\right)\left(b+2c\right)}}=\dfrac{a}{3}\)

Tương tự:

\(\dfrac{b^3}{\left(b+2c\right)\left(c+2a\right)}+\dfrac{b+2c}{27}+\dfrac{c+2a}{27}\ge\dfrac{b}{3}\)

\(\dfrac{c^3}{\left(c+2a\right)\left(a+2b\right)}+\dfrac{c+2a}{27}+\dfrac{a+2b}{27}\ge\dfrac{c}{3}\)

Cộng vế:

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

\(\Rightarrow VT\ge\dfrac{a+b+c}{9}\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c\)

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

\(P\ge\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)}=\dfrac{\left(a^2+b^2+c^2+3abc\right)^2}{\left(ab+bc+ca\right)^2}\)

\(P\ge\dfrac{\left[a^2+b^2+c^2+3abc\right]^2}{\left(ab+bc+ca\right)^2}\)

Do đó ta chỉ cần chứng minh \(\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge2\)

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

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

\(\Leftrightarrow3abc\ge4\left(ab+bc+ca\right)-9\)

\(\Rightarrow\dfrac{a^2+b^2+c^2+3abc}{ab+bc+ca}\ge\dfrac{a^2+b^2+c^2+4\left(ab+bc+ca\right)-9}{ab+bc+ca}\)

\(=\dfrac{\left(a+b+c\right)^2-9+2\left(ab+bc+ca\right)}{ab+bc+ca}=2\) (đpcm)

sai cơ bản rồi bạn ơi : a(a+bc)^2 không bằng dc (a^2+abc)^2

a)ta có \(a^2+b^2\ge2ab\)

\(\Leftrightarrow1\ge ab\)

theo bđt cauchy schwarz ta có

\(\left(\dfrac{a}{b}+\dfrac{b}{a}\right)\left(\dfrac{a}{b^2}+\dfrac{b}{a^2}\right)\ge2\sqrt{\dfrac{a.b}{a.b}}.2\sqrt{\dfrac{a.b}{a^2.b^2}}=2.1.2\dfrac{1}{1^2}=4\)

\(\Rightarrow dpcm\)

I. Đúng do BĐT Cosi \(a+\dfrac{9}{a}\ge2.\sqrt{a.\dfrac{9}{a}}=6\)

II. Sai do \(\dfrac{a^2+5}{\sqrt{a^2+4}}=\sqrt{a^2+4}+\dfrac{1}{\sqrt{a^2+4}}\ge2+\dfrac{1}{a^2+4}>2\)

III. Đúng do BĐT Cosi \(\dfrac{\sqrt{ab}}{ab+1}\le\dfrac{\sqrt{ab}}{2\sqrt{ab}}=\dfrac{1}{2}\)

IV. Đúng do BĐT BSC \(\left(a+\dfrac{1}{b}\right)\left(b+\dfrac{1}{a}\right)\ge\left(\sqrt{a}.\dfrac{1}{\sqrt{a}}+\sqrt{b}.\dfrac{1}{\sqrt{b}}\right)^2=4\)