Bài 1:

Áp dụng BĐT AM-GM ta có:

$\frac{1}{a+1}+\frac{1}{b+1}+\frac{1}{c+1}\geq 3\sqrt[3]{\frac{1}{(a+1)(b+1)(c+1)}}$

$\frac{a}{a+1}+\frac{b}{b+1}+\frac{c}{c+1}\geq 3\sqrt[3]{\frac{abc}{(a+1)(b+1)(c+1)}}$

Cộng theo vế và thu gọn:

$\frac{a+1}{a+1}+\frac{b+1}{b+1}+\frac{c+1}{c+1}\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Leftrightarrow 3\geq \frac{3(1+\sqrt[3]{abc})}{\sqrt[3]{(a+1)(b+1)(c+1)}}$

$\Rightarrow (a+1)(b+1)(c+1)\geq (1+\sqrt[3]{abc})^3$

Ta có đpcm.

Bài 2:

$a^3+a^3+a^3+a^3+b^3+c^3\geq 6\sqrt[6]{a^{12}b^3c^3}=6a^2\sqrt{bc}$

$b^3+b^3+b^3+b^3+a^3+c^3\geq 6b^2\sqrt{ac}$

$c^3+c^3+c^3+c^3+a^3+b^3\geq 6c^2\sqrt{ab}$

Cộng theo vế và rút gọn thu được:

$a^3+b^3+c^3\geq a^2\sqrt{bc}+b^2\sqrt{ac}+c^2\sqrt{ab}$ 

Ta có đpcm.

Dấu "=" xảy ra khi $a=b=c$

Bài 3:

Áp dụng BĐT Cauchy-Schwarz:

$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\geq \frac{(a+b+c)^2}{b+c+c+a+a+b}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$

\(\Leftrightarrow\left(1+abc\right)\left(\dfrac{1}{a\left(1+b\right)}+\dfrac{1}{b\left(1+c\right)}+\dfrac{1}{c\left(1+a\right)}\right)\ge3\)

Ta có:

\(\left(1+abc\right).\dfrac{1}{a\left(1+b\right)}=\dfrac{1+abc}{a+ab}=\dfrac{1+a+ab+abc-a-ab}{a+ab}=\dfrac{1+a}{a\left(1+b\right)}+\dfrac{b\left(1+c\right)}{1+b}-1\)

\(\Rightarrow VT=\dfrac{1+a}{a\left(1+b\right)}+\dfrac{b\left(1+c\right)}{1+b}+\dfrac{1+b}{b\left(1+c\right)}+\dfrac{c\left(1+a\right)}{1+c}+\dfrac{1+c}{c\left(1+a\right)}+\dfrac{a\left(1+b\right)}{1+a}-3\)

\(VT\ge6\sqrt[6]{\dfrac{abc\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}{abc\left(1+a\right)^2\left(1+b\right)^2\left(1+c\right)^2}}-3=3\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Ta chứng minh 2 bất đẳng thức phụ sau: với x, y, z dương thì:

\(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\left(1\right)\)

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)\ge\left(1+\sqrt[3]{xyz}\right)^3\left(2\right)\)

+ Chứng minh BĐT (1), sử dụng BĐT AM - GM:

\(x^4+x^4+y^4+z^4\ge4x^2yz\)

\(y^4+y^4+x^4+z^4\ge4xy^2z\)

\(z^4+z^4+x^4+y^4\ge4xyz^2\)

Cộng dồn lại ta có: \(x^4+y^4+z^4\ge xyz\left(x+y+z\right)\)

+ Chứng minh BĐT (2). Ta có:

\(\left(1+x\right)\left(1+y\right)\left(1+z\right)=1+x+y+z+xy+yz+xyz\ge1+3\sqrt[3]{xyz}+3\sqrt[3]{x^2y^2z^2}+xyz=\left(1+\sqrt[3]{xyz}\right)^3\)

Bây giờ ta quay lại chứng minh BĐT ở đề.

BĐT cần chứng minh tương đương với BĐT sau:

\(\sqrt[4]{\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4}\ge\sqrt[4]{3}+\dfrac{\sqrt[4]{243}}{2+abc}\)

\(\Leftrightarrow\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Sử dụng BĐT (1) ta có:

\(\left(1+\dfrac{1}{a}\right)^4+\left(1+\dfrac{1}{b}\right)^4+\left(1+\dfrac{1}{c}\right)^4\ge\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

Sử dụng BĐT (2) và BĐT AM - GM ta có:

\(\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\left(1+\dfrac{1}{\sqrt[3]{abc}}\right)^3\left(3+\dfrac{3}{\sqrt[3]{abc}}\right)\)

\(\Rightarrow\left(1+\dfrac{1}{a}\right)\left(1+\dfrac{1}{b}\right)\left(1+\dfrac{1}{c}\right)\left(3+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge3\left(1+\dfrac{1}{\sqrt[3]{abc.1.1}}\right)^4\ge3\left(1+\dfrac{3}{2+abc}\right)^4\)

Vậy BĐT đã được chứng minh. Đẳng thức xảy ra <=> a = b = c.

Áp dụng BĐT: \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\left(\text{luôn đúng}\right)\)

Ta có \(\dfrac{a}{a+1}+\dfrac{b}{b+1}+\dfrac{c}{c+1}\ge3\sqrt[3]{\dfrac{abc}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

\(\dfrac{1}{a+1}+\dfrac{1}{b+1}+\dfrac{1}{c+1}\ge3\sqrt[3]{\dfrac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\)

Cộng VTV \(\Leftrightarrow3\ge\dfrac{3\left(\sqrt[3]{abc}+1\right)}{\sqrt[3]{\left(a+1\right)\left(b+1\right)\left(c+1\right)}}\Leftrightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge\sqrt[3]{abc}+1\)

\(\Leftrightarrow VT^2=\sum\left[\dfrac{1}{a\left(b+1\right)}\right]^2\ge3\cdot\sum\dfrac{1}{ab\left(a+1\right)\left(b+1\right)}\\ \Leftrightarrow VT^2\ge3\cdot\dfrac{a^2+b^2+c^2+a+b+c}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge3\cdot\dfrac{a+b+c+ab+bc+ca}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\\ \Leftrightarrow VT^2\ge\dfrac{3}{abc}-\dfrac{3\left(abc+1\right)}{abc\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{3}{abc}-\dfrac{3\left(abc+1\right)}{abc\left(1+\sqrt[3]{abc}\right)^3}\\ \Leftrightarrow VT^2\ge\dfrac{9}{\sqrt[3]{\left(abc\right)^2}\left(1+\sqrt[3]{abc}\right)^2}=VP^2\\ \LeftrightarrowĐpcm\)

Dấu \("="\Leftrightarrow a=b=c=1\)

Áp dụng BĐT AM - GM ta có:

$ \frac{a^3}{(1 + b)(1 + c)} + \frac{1 + b}{8} + \frac{1 + c}{8} \geq \frac{3}{4}a$

$\frac{b^3}{(1 + c)(1 + a)} + \frac{1 + c}{8} + \frac{1 + a}{8} \geq \frac{3}{4}b$

$\frac{c^3}{(1 + a)(1 + b)} + \frac{1 + a}{8} + \frac{1 + b}{8} \geq \frac{3}{4}c $

Cộng vế theo vế ta được:

$ P + \frac{2(a + b + c) + 6}{8} \geq \frac{3}{4}(a + b + c) $

$<=> P \geq \frac{1}{2}(a + b + c) - \frac{3}{4}$

$=> P \geq \frac{3}{4} (dpcm)$

Giải:

\(\dfrac{a}{\left(a+1\right)\left(b+1\right)}+\dfrac{b}{\left(b+1\right)\left(c+1\right)}+\dfrac{c}{\left(c+1\right)\left(a+1\right)}\ge\dfrac{3}{4}\)(*)

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

\(\Leftrightarrow\) \(\dfrac{ac+a+ab+b+bc+c}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\) \(\ge\) \(\dfrac{3}{4}\)

Do a+1 ; b+1; c+1 >0

\(\Rightarrow\) 4ac+4a+4ab+4b+4bc+4c \(\ge\) 3abc+3ac+3bc+3ab+3a+3b+3c+3

\(\Leftrightarrow\) ac+ab+bc+a+b+c -6 \(\ge\) 0

Áp dụng BĐT Cô-si cho 3 số

Ta có: a+b+c \(\ge\) \(3\sqrt[3]{abc}=3\)

ab+bc+ca \(\ge\) \(3\sqrt[3]{\left(abc\right)^2}\) = 3

\(\Rightarrow\)ac+ab+bc+a+b+c -6 \(\ge\) 0 ( luôn đúng)

\(\Rightarrow\) (*) được chứng minh

Dấu "=" xảy ra \(\Leftrightarrow\) a=b=c=1

Lời giải:

Ta có:

\(\text{VT}=\frac{a}{(a+1)(b+1)}+\frac{b}{(b+1)(c+1)}+\frac{c}{(c+1)(a+1)}\)

\(=\frac{a(c+1)+b(a+1)+c(b+1)}{(a+1)(b+1)(c+1)}=\frac{ab+bc+ac+a+b+c}{abc+(ab+bc+ac)+(a+b+c)+1}\)

\(=\frac{ab+bc+ac+a+b+c}{2+(a+b+c)+ab+bc+ac}\)

Ta cần chứng minh \(\text{VT}\geq \frac{3}{4}\)

\(\Leftrightarrow \frac{ab+bc+ac+a+b+c}{2+(a+b+c)+ab+bc+ac}\geq \frac{3}{4}\)

\(\Leftrightarrow 4(ab+bc+ac+a+b+c)\geq 3(ab+bc+ac+a+b+c)+6\)

\(\Leftrightarrow ab+bc+ac+a+b+c\geq 6\)

\(\Leftrightarrow ab+bc+ac+a+b+c\geq 6\sqrt[6]{ab.bc.ac.a.b.c}\)

(Đúng theo BĐT Cô-si)

Do đó ta có đpcm

Dấu bằng xảy ra khi \(a=b=c=1\)

điện thoại cùi nên chụp hơi mờ, đề này còn thiếu a,,bc>0

Đề bài hình như bị sai em, thay điểm rơi ko thỏa mãn

Biểu thức là \(a+b+\sqrt{2\left(a+c\right)}\) mới đúng

em cũng nghĩ thế mới dùng đc BDT AM-GM 3 số đúng ko thầy :)