\(\frac{a}{3}=\frac{3}{b}\Rightarrow ab=9\)

\(\frac{a}{3}=\frac{a}{b}\Rightarrow ab=3a\)

=> 3a =9 => a =3

a=3 => 3.b=9  => b=3

Vậy a =b =3

Áp dụng bđt cosi ta có VT = a+1/b.(a-b) = (a-b) + 1/b.(a-b) + b >= 3.\(\sqrt[3]{\frac{\left(a-b\right).1.b}{\left(a-b\right).b}}\)=3

=> ĐPCM 

Ta có:

\(A=2a+\frac{32}{\left(a-b\right)\left(2b+3\right)^2}\)

\(=\frac{2b+3}{2}+\frac{2b+3}{2}+2\left(a-b\right)+\frac{32}{\left(a-b\right)\left(2b+3\right)^2}-3\)

Theo BĐT cô-si ta có:

\(A\ge4\sqrt[4]{\frac{2b+3}{2}.\frac{2b+3}{2}.2\left(a-b\right).\frac{32}{\left(a-b\right)\left(2b+3\right)^2}}-3\)

\(\Leftrightarrow A\ge4\sqrt[4]{16}-3=5\)

=> ĐPCM

hok tốt

\(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

\(\frac{\left(a+b\right)^3}{c^3}+\frac{\left(b+c\right)^3}{a^3}+\frac{\left(c+a\right)^3}{b^3}\)

\(\frac{\left(a^2b+ab^2\right)^3+\left(b^2c+c^2b\right)^3+\left(c^2a+a^2c\right)^3}{\left(abc\right)^3}\)

\(\left(\frac{a}{c}+\frac{b}{c}\right)^3+\left(\frac{b}{a}+\frac{c}{a}\right)^3+\left(\frac{c}{b}+\frac{a}{b}\right)^3\)

\(\left(\frac{a+b}{c}\right)^3+\left(\frac{b+c}{a}\right)^3+\left(\frac{c+a}{b}\right)^3\)

dễ thấy \(\frac{c}{a+b}=\frac{1}{2}< =>\frac{a+b}{c}=2\)

làm tương tự với 3 cái còn lại ta đc:

\(2^3+2^3+2^3=24\)

Ta có:a/c=c/b=>c²=ab

thay vào biểu thức ta có:

VT=a²+c²/b²+c²=a²+ab/b²+ab=a(a+b)/b(a+b)=a/b

 Vì VT=VP(=a/b)

=>đpcm