Question

[vũ văn tùng]

Cho các số thực a;b;c > 1 . Chứng minh rằng  1/(2a - 1)  + 1/(2b - 1)  + 1/(2c - 1)   + 4ab/(1 + ab) + 4bc/(1 + bc)  + 4ca/(1 + ca)  > 9

Answer 1

[zZz Cool Kid_new zZz]

Do \(a,b,c\ge1\) nên \(\left(a-1\right)\left(b-1\right)\ge0\Leftrightarrow ab+1\ge a+b\)

Mà \(\frac{4ab}{1+ab}=\frac{4\left(1+ab\right)-4}{1+ab}=4-\frac{4}{1+ab}\ge4-\frac{4}{a+b}\)

Tương tự:\(\frac{4bc}{1+bc}\ge4-\frac{4}{b+c};\frac{4ca}{1+ca}\ge4-\frac{4}{c+a}\)

Mặt khác:\(\left(a-1\right)^2\ge0\Leftrightarrow a^2\ge2a-1\)

Tương tự:\(b^2\ge2b-1;c^2\ge2c-1\)

Khi đó ta có:

\(LHS\ge\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+12-4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)=S\)

Áp dụng AM - GM ta dễ có:\(\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\ge\frac{2}{\frac{\left(a+b\right)^2}{4}}=\frac{8}{\left(a+b\right)^2}\)

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

\(\Rightarrow S\ge\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}+12-4\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\)

\(=\left(\frac{2}{a+b}-1\right)^2+\left(\frac{2}{b+c}-1\right)^2+\left(\frac{2}{c+a}-1\right)^2+9\)

\(\ge9\)

 Vậy ta có đpcm

Đẳng thức xảy ra tại \(a=b=c=1\)

Answer 2

[vũ văn tùng]

Mình cảm ơn bạn