-Ta c/m: Với a,b,c là các số thực thỏa mãn \(abc=1\) thì:
\(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}=1\)
-Ta có:
\(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\)
\(=\dfrac{1}{\dfrac{1}{c}+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{\dfrac{1}{b}+\dfrac{1}{bc}+1}\)
\(=\dfrac{1}{\dfrac{1+bc+c}{c}}+\dfrac{1}{bc+c+1}+\dfrac{1}{\dfrac{c+1+bc}{bc}}\)
\(=\dfrac{c}{bc+c+1}+\dfrac{1}{bc+c+1}+\dfrac{bc}{bc+c+1}=\dfrac{bc+c+1}{bc+c+1}=1\) (đpcm).
-Quay lại bài toán, nhưng với các số thực a,b,c dương:
\(A=\dfrac{1}{a^2+2b^2+3}+\dfrac{1}{b^2+2c^2+3}+\dfrac{1}{c^2+2a^2+3}\)
\(=\dfrac{1}{a^2+b^2+b^2+1+2}+\dfrac{1}{b^2+c^2+c^2+1+2}+\dfrac{1}{c^2+a^2+a^2+1+2}\)
-Áp dụng BĐT AM-GM ta có:
\(A=\dfrac{1}{a^2+b^2+b^2+1+2}+\dfrac{1}{b^2+c^2+c^2+1+2}+\dfrac{1}{c^2+a^2+a^2+1+2}\le\dfrac{1}{2ab+2b+2}+\dfrac{1}{2bc+2c+2}+\dfrac{1}{2ca+2a+2}=\dfrac{1}{2}\left(\dfrac{1}{ab+b+1}+\dfrac{1}{bc+c+1}+\dfrac{1}{ca+a+1}\right)=\dfrac{1}{2}\)-Dấu "=" xảy ra khi \(a=b=c=1\)
