Cách ngắn gọn:

\(1+\frac{1}{a^3}=\frac{1}{8}+\frac{1}{8}+...+\frac{1}{a^3}\ge9\sqrt[9]{\frac{1}{8^8.a^3}}=9\sqrt[9]{\frac{1}{8^8}}.\sqrt[3]{\frac{1}{a}}\)

Tương tự với b, c

\(\Rightarrow\left(1+\frac{1}{a^3}\right)\left(1+\frac{1}{b^3}\right)\left(1+\frac{1}{c^3}\right)\ge\left(9\sqrt[9]{\frac{1}{8^8}}\right)^3.\sqrt[3]{\frac{1}{abc}}\ge\frac{729}{256}.\sqrt[3]{\frac{1}{\left(\frac{a+b+c}{3}\right)^3}}=\frac{729}{512}\)

Dấu "=" xảy ra khi a = b = c = 2.

BĐT\(\Leftrightarrow\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(a+c\right)}+\frac{abc}{c^3\left(a+b\right)}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Leftrightarrow\frac{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}{\frac{1}{b}+\frac{1}{c}.\frac{1}{a}+\frac{1}{c}.\frac{1}{a}+\frac{1}{b}}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Đặt \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\). Áp dụng BĐT: AM-GM ta có:

\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}}=a\)

\(\frac{b^2}{a+b}+\frac{a+c}{4}\ge2\sqrt{\frac{b^2}{a+b}.\frac{a+b}{4}}=b\)

\(\frac{c^2}{a+b}+\frac{a+b}{4}\ge2\sqrt{\frac{c^2}{a+b}+\frac{a+b}{4}}=c\)

Cộng theo vế 3 BĐT trên ta có:

\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

hay \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{3}{2}\)

Dấu bằng = xảy ra khi a = b = c = 1

Đặt  \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\Rightarrow xyz=1;x>0;y>0;z>0\)

Ta cần chứng minh bất đẳng thức sau : \(A=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{3}{2}\)

Sử dụng bất đẳng thức Bunhiacopxki cho 2 bộ số :

\(\left(\sqrt{y+z};\sqrt{z+x};\sqrt{x+y}\right);\left(\frac{x}{\sqrt{y+z}};\frac{y}{\sqrt{z+x}};\frac{z}{\sqrt{x+y}}\right)\)

Ta có : \(\left(x+y+z\right)^2\le\left(x+y+z+x+y+z\right)A\)

\(\Rightarrow A\ge\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\left(Q.E.D\right)\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z=1\Leftrightarrow a=b=c=1\)

Bạn tham khảo tại đây:

Câu hỏi của Trần Hữu Ngọc Minh - Toán lớp 9 - Học toán với OnlineMath

Áp dụng BĐT Cosi ta được:

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge3\sqrt{\frac{a^3\left(1+b\right)\left(1+c\right)}{\left(1+b\right)\left(1+c\right)64}}=\frac{3a}{4}̸\)

Tương tự \(\hept{\begin{cases}\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{1+a}{8}+\frac{1+c}{8}\ge\frac{3b}{4}\\\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{1+a}{8}+\frac{1+b}{8}\ge\frac{3c}{4}\end{cases}}\)

Cộng theo từng vế BĐT trên ta có:

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{3}{4}\ge\frac{a+b+c}{2}\)

Vì \(a+b+c\ge3\sqrt[3]{abc}=3\)do đó:

\(\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{b^3}{\left(1+a\right)\left(1+c\right)}+\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{3}{4}\ge\frac{3}{2}\)

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

Đẳng thức xảy ra <=> a=b=c

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow\frac{a^3}{\left(1+b\right)\left(1+c\right)}+\frac{1+b}{8}+\frac{1+c}{8}\ge3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)

Tượng tự ta có \(\hept{\begin{cases}\frac{b^3}{\left(1+c\right)\left(1+a\right)}+\frac{1+c}{8}+\frac{1+a}{8}\ge\frac{3b}{4}\\\frac{c^3}{\left(1+a\right)\left(1+b\right)}+\frac{1+a}{8}+\frac{1+b}{8}\ge\frac{3c}{4}\end{cases}}\)

\(\Rightarrow VT+\frac{3}{4}+\frac{a+b+c}{4}\ge\frac{3\left(a+b+c\right)}{4}\)

\(\Rightarrow VT\ge\frac{a+b+c}{2}-\frac{3}{4}\)(1) 

Áp dụng bất đẳng thức Cauchy - Schwarz 

\(\Rightarrow a+b+c\ge3\sqrt[3]{abc}=3\)

\(\Rightarrow\frac{a+b+c}{2}-\frac{3}{4}\ge\frac{3}{4}\)(2) 

Từ (1) và (2) 

\(\Rightarrow VT\ge\frac{3}{4}\)( đpcm ) 

Dấu " = " xảy ra khi \(a=b=c=1\)

Áp dụng bất đẳng thức AM-GM cho 3 số :

\(\frac{a^3}{\left(b+1\right)\left(c+1\right)}+\frac{b+1}{8}+\frac{c+1}{8}\ge3\sqrt[3]{\frac{a^3\left(b+1\right)\left(c+1\right)}{\left(b+1\right)\left(c+1\right)8^2}}=\frac{3a}{4}\)

Tương tự ta có \(\frac{b^3}{\left(c+1\right)\left(a+1\right)}+\frac{c+1}{8}+\frac{a+1}{8}\ge\frac{3b}{4}\)

\(\frac{c^3}{\left(a+1\right)\left(b+1\right)}+\frac{a+1}{8}+\frac{b+1}{8}\ge\frac{3c}{4}\)

Cộng theo vế các bđt trên ta được : 

\(VT+2\left(\frac{a}{8}+\frac{b}{8}+\frac{c}{8}+\frac{3}{8}\right)\ge\frac{3}{4}\left(a+b+c\right)\)

\(< =>VT\ge\frac{3}{4}\left(a+b+c\right)-\frac{1}{4}\left(a+b+c\right)-\frac{6}{8}\)

\(=\frac{1}{2}\left(a+b+c\right)-\frac{6}{8}\ge\frac{1}{2}.3\sqrt[3]{abc}-\frac{6}{8}=\frac{12-6}{8}=\frac{6}{8}=\frac{3}{4}\)

Dấu "=" xảy ra \(< =>a=b=c=1\)

Done !