Lời giải:

Áp dụng BĐT AM-GM ta có:

$\text{VT}=[\frac{a+1}{4}+\frac{1}{a+1}+\frac{3}{4}a-\frac{1}{4}][\frac{b+1}{4}+\frac{1}{b+1}+\frac{3}{4}b-\frac{1}{4}][\frac{c+1}{4}+\frac{1}{c+1}+\frac{3}{4}c-\frac{1}{4}]$

$\geq [2\sqrt{\frac{1}{4}}+\frac{3}{4}a-\frac{1}{4}][2\sqrt{\frac{1}{4}}+\frac{3}{4}b-\frac{1}{4}][2\sqrt{\frac{1}{4}}+\frac{3}{4}c-\frac{1}{4}]$
$=\frac{3}{4}(a+1).\frac{3}{4}(b+1).\frac{3}{4}(c+1)$
$=\frac{27}{64}(a+1)(b+1)(c+1)$

$\geq \frac{27}{64}.2\sqrt{a}.2\sqrt{b}.2\sqrt{c}$

$=\frac{27}{64}.8\sqrt{abc}\geq \frac{27}{64}.8=\frac{27}{8}$ (đpcm)

Dấu "=" xảy ra khi $a=b=c=1$

\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)

\(\dfrac{1}{1+a^2\left(b+c\right)}=\dfrac{1}{1+a\left(ab+ac\right)}=\dfrac{1}{1+a\left(3-bc\right)}=\dfrac{1}{1+3a-abc}=\dfrac{1}{3a+\left(1-abc\right)}\le\dfrac{1}{3a}\)

Tương tự và cộng lại:

\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)

Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có ít nhất 2 số cùng phía so với 1

Không mất tính tổng quát, giả sử đó là a và b

\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)

\(\Leftrightarrow ab+1\ge a+b\)

\(\Leftrightarrow2\left(ab+1\right)\ge\left(a+1\right)\left(b+1\right)\)

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

Lại có:

\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1\right)^2}\ge\dfrac{1}{\left(ab+1\right)\left(\dfrac{a}{b}+1\right)}+\dfrac{1}{\left(ab+1\right)\left(\dfrac{b}{a}+1\right)}=\dfrac{1}{ab+1}\)

\(\Rightarrow P\ge\dfrac{1}{ab+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}=\dfrac{1}{\dfrac{1}{c}+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}\)

\(\Rightarrow P\ge\dfrac{c}{c+1}+\dfrac{c+1}{\left(c+1\right)^2}=\dfrac{c\left(c+1\right)+c+1}{\left(c+1\right)^2}=\dfrac{\left(c+1\right)^2}{\left(c+1\right)^2}=1\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Lời giải:

Đặt biểu thức vế trái là A

Có \(a+\frac{1}{a+1}=\frac{a^2+a+1}{a+1}=\frac{a^2}{a+1}+1=\frac{a^2}{a+1}+\frac{1}{2}+\frac{1}{2}\)

Áp dụng BĐT Cauchy-Schwarz:
\(a+\frac{1}{a+1}\geq \frac{(a+1+1)^2}{a+1+2+2}=\frac{(a+2)^2}{a+5}\)

Thực hiện tương tự với các phân thức còn lại và nhân theo vế:

\(\Rightarrow A\geq \frac{(a+2)^2(b+2)^2(c+2)^2}{(a+5)(b+5)(c+5)}\)

Áp dụng BĐT AM-GM:

\((a+2)(b+2)(c+2)\geq 3\sqrt[3]{a}.3\sqrt[3]{b}.3\sqrt[3]{c}=27\sqrt[3]{abc}\geq 27\)

\(\Rightarrow A\geq \frac{27(a+2)(b+2)(c+2)}{(a+5)(b+5)(c+5)}\) (1)

Ta sẽ cm

\(\frac{27(a+2)(b+2)(c+2)}{(a+5)(b+5)(c+5)}\geq \frac{27}{8}(*)\Leftrightarrow 8(a+2)(b+2)(c+2)\geq (a+5)(b+5)(c+5)\)

\(\Leftrightarrow 8[abc+8+2(ab+bc+ac)+4(a+b+c)]\geq abc+125+5(ab+bc+ac)+25(a+b+c)\)

\(\Leftrightarrow 7abc+11(ab+bc+ac)+7(a+b+c)\geq 61\)

BĐT trên luôn đúng theo AM_GM:

\(7abc+11(ab+bc+ac)+7(a+b+c)\geq 7abc+33\sqrt[3]{a^2b^2c^2}+21\sqrt[3]{abc}\geq 7+33+21=61\)

Do đó (*) đúng.

Từ \((1);(2)\Rightarrow A\geq \frac{27}{8}\) (đpcm)

Dấu bằng xảy ra khi \(a=b=c=1\)

Ta có:

\(\dfrac{1}{a^3\left(b+c\right)}+\dfrac{1}{b^3\left(c+a\right)}+\dfrac{1}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)

\(\dfrac{1^2}{a^3\left(b+c\right)}+\dfrac{1^2}{b^3\left(c+a\right)}+\dfrac{1^2}{c^3\left(a+b\right)}\ge\dfrac{3}{2}\)

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

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

Áp dụng BĐT Svacxo ta có:

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

Chứng minh: \(\dfrac{ab+bc+ca}{2}\ge\dfrac{3}{2}\Leftrightarrow ab+bc+ca\ge3\)

Áp dụng BĐT Cosi ta có:

\(ab+bc+ca\ge3\sqrt[3]{ab.bc.ca}\)

\(ab+bc+ca\ge3\) (2)

Từ (1) và (2)

=> ĐPCM

chết đăng nhầm sogy nha

Tách biểu thức như sau:

\(\left(\dfrac{a}{9}+\dfrac{b}{12}+\dfrac{c}{6}+\dfrac{8}{abc}\right)+\left(\dfrac{a}{18}+\dfrac{b}{24}+\dfrac{2}{ab}\right)+\left(\dfrac{b}{16}+\dfrac{c}{8}+\dfrac{2}{bc}\right)+\left(\dfrac{a}{9}+\dfrac{c}{6}+\dfrac{2}{ca}\right)+\left(\dfrac{13a}{18}+\dfrac{13b}{24}\right)+\left(\dfrac{13b}{48}+\dfrac{13c}{24}\right)\)

ta có \(\dfrac{1}{\left(a+b\right)c}\le\dfrac{1}{2\sqrt{ab}c}=\dfrac{1}{2\sqrt{c}}\)tương tự ta có

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

Đặt \(a\left(1-b\right)=x;b\left(1-c\right)=y;c\left(1-a\right)=x\)

\(\Rightarrow1-\left(a+b+c\right)+ab+bc+ca=1-a\left(1-b\right)-b\left(1-c\right)-c\left(1-a\right)=1-x-y-z\)

BĐT cần c/m trở thành:

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{3}{1-x-y-z}\)

\(\Leftrightarrow\left(1-x-y-z\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-3\ge0\)

\(\Leftrightarrow\dfrac{1-x-y-z}{x}+\dfrac{1-x-y-z}{y}+\dfrac{1-x-y-z}{z}-3\ge0\)

\(\Leftrightarrow\dfrac{1-y-z}{x}+\dfrac{1-z-x}{y}+\dfrac{1-x-y}{z}-6\ge0\) (1)

Lại có: \(1-y-z=1-b\left(1-c\right)-c\left(1-a\right)=1-b-c+bc+ca=\left(1-b\right)\left(1-c\right)+ca\)

Nên (1) tương đương:

\(\dfrac{\left(1-b\right)\left(1-c\right)+ca}{a\left(1-b\right)}+\dfrac{\left(1-a\right)\left(1-c\right)+ab}{b\left(1-c\right)}+\dfrac{\left(1-a\right)\left(1-b\right)+bc}{c\left(1-a\right)}-6\ge0\)

\(\Leftrightarrow\dfrac{1-c}{a}+\dfrac{c}{1-b}+\dfrac{1-a}{b}+\dfrac{a}{1-c}+\dfrac{1-b}{c}+\dfrac{b}{1-a}\ge6\)

BĐT trên hiển nhiên đúng theo AM-GM do:

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

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{2}\)