Question

Cho 3 số thực dương a, b, c. CMR: \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\) thì \(\frac{\left(a+b\right)^3}{c^3}+\frac{\left(b+c\right)^3}{a^3}+\frac{\left(c+a\right)^3}{b^3}=24\)

Answer 1

 a + b + c = 3 

a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥ 3/2 

Ta có 

a/( 1 + b^2 ) = a - ab^2/( 1 + b^2 ) ≥ a - ab^2/2b = a - ab/2 

Tương tự ta có 

b/( 1 + c^2 ) ≥ b - bc/2 

c/( 1 + a^2 ) ≥ c - ac/2 

Cộng vào ta có 

a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥ a + b + c - ( ab + bc + ac )/2 = 3 - ( ab + bc + ac )/2 

Xét ab + bc + ac 

Ta có 

a^2 + b^2 ≥ 2ab 
b^2 + c^2 ≥ 2bc 
c^2 + a^2 ≥ 2ac 

=> a^2 + b^2 + c^2 ≥ ab + bc + ac 

<=> a^2 + b^2 + c^2 + 2ac + 2bc + 2ab ≥ 3( ab + ac + bc ) 

<=> ( a + b + c )^2 ≥ 3( ab + ac + bc ) 

<=> ab + ac + bc ≤ 9:3 = 3 

=> 3 - ( ab + bc + ac )/2 ≥ 3 - 3/2 = 3/2 

=> a/( 1 + b^2 ) + b/( 1 + c^2 ) + c/( 1 + a^2 ) ≥   

Answer 2

Theo dãy tỉ số bằng nhau , có :

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

\(\Rightarrow\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}=2\)

\(\Rightarrow\left(\frac{b+c}{a}\right)^3=\left(\frac{c+a}{b}\right)^3=\left(\frac{a+b}{c}\right)^3=8\)

\(\Rightarrow\frac{\left(b+c\right)^3}{a^3}+\frac{\left(c+a\right)^3}{b^3}+\frac{\left(a+b\right)^3}{c^3}=8.3=24\)