ĐK:3x+2>=0

  <=>3x>=-2

 <=> x>=-2/3

Trả lời:

Cần CM: \(\frac{1}{9-a}-\frac{12}{a^2+63}\ge\frac{1}{144}a^2-\frac{1}{16}\) (1) 

\(\Leftrightarrow\)\(\frac{a^2+12a-45}{\left(9-a\right)\left(a^2+63\right)}\ge\frac{1}{144}a^2-\frac{1}{16}\)

\(\Leftrightarrow\)\(144\left(a^2+12a-45\right)\ge\left(a-3\right)\left(a+3\right)\left(9-a\right)\left(a^2+63\right)\)

\(\Leftrightarrow\)\(\left(a-3\right)\left[144\left(a+15\right)-\left(a+3\right)\left(9-a\right)\left(a^2+63\right)\right]\ge0\)

\(\Leftrightarrow\)\(\left(a-3\right)\left(a^4-6a^3+36a^2-234a+459\right)\ge0\)

\(\Leftrightarrow\)\(\left(a-3\right)^2\left(a^3-3a^2+27a+153\right)\ge0\)

\(\Leftrightarrow\)\(\left(a-3\right)^2\left[\left(a-3\right)^2\left(a+3\right)+36a+126\right]\ge0\) ( đúng )

Do đó (1) đúng => \(\Sigma_{cyc}\frac{1}{9-a}-\Sigma_{cyc}\frac{12}{a^2+63}\ge\frac{1}{144}\left(a^2+b^2+c^2\right)-\frac{3}{16}=0\)

\(\Rightarrow\)\(\Sigma_{cyc}\frac{12}{a^2+63}\le\Sigma_{cyc}\frac{1}{9-a}\le\Sigma_{cyc}\frac{1}{a+b}\) ( do \(a+b+c\le9\) ) 

Dấu "=" xảy ra khi a=b=c=3 

Xét a = b = c = 1 thì thỏa mãn bài ra

Xét a ,b,c khác 1. do a,b,c có vai trò như nhau nên giả sử \(a\le b\le c\)

Áp dụng BĐT cô-si cho 3 số a+b+1,1-a,1-b, ta có :

\(\left(a+b+1\right)\left(1-a\right)\left(1-b\right)\le\left(\frac{a+b+1+1-a+1-b}{3}\right)^3=1\)

\(\Rightarrow\left(1-a\right)\left(1-b\right)\le\frac{1}{a+b+1}\)

\(\Rightarrow\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\frac{1-c}{a+b+1}\)

Mà \(\frac{a}{b+c+1}\le\frac{a}{a+b+1};\frac{b}{a+c+1}\le\frac{b}{a+b+1}\)

\(\Rightarrow\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}\le\frac{a}{a+b+1}+\frac{b}{a+b+1}+\frac{c}{a+b+1}\)

do đó : \(\frac{a}{b+c+1}+\frac{b}{a+c+1}+\frac{c}{a+b+1}+\left(1-a\right)\left(1-b\right)\left(1-c\right)\)

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

dấu " = " xảy ra khi a = b = c = 0

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