Cho các số thực dương [TEX]a,b,c [/TEX]sao cho [TEX]a+b+c=3[/TEX]. Chứng minh rằng
[TEX]\frac{1}{9-bc}+\frac{1}{9-ac}+\frac{1}{9-ab} \le \frac{3}{8}[/TEX]
[tex]Dat ab=x,ac=y,bc=z[/tex]
\Leftrightarrow[TEX]\frac{1-x}{9-x}+\frac{1-y}{9-y}+\frac{1-z}{9-z} \ge 0[/TEX]
[TEX]\frac{(1-x)(x+6)}{(9-x)(x+6)}+\frac{(1-y)(y+6)}{(9-y)(y+6)}+\frac{(1-z)(z+6)}{(9-z)(z+6)} \ge 0[/TEX]
[tex]By ---Chebyshev---[/tex]
cho 2 dãy cùng chiều :
[tex](1-x)(x+6),(1-y)(y+6),(1-z)(z+6)[/tex]
[tex]\frac{1}{(9-x)(x+6)},\frac{1}{(9-y)(y+6)},\frac{1}{(9-z)(z+6)},[/tex]
[tex]VT \ge \sum[(1-x)(x+6)].\sum{\frac{1}{(9-x)(x+6)}}[/tex]
Vậy chỉ cần CM:
[tex]\sum[(1-x)(x+6)=18-5(x+y+z)-(x^2+y^2+z^2) \ge 0[/tex]
[tex]\Leftrightarrow 5(ab+bc+ac)+(a^2b^2+b^2c^2+a^2c^2) \le 18 [/tex]
[tex]\Leftrightarrow 5(ab+bc+ac)+(ab+bc+ac)^2 \le 18+6abc(1)[/tex]
Theo CHUR:
[tex](a+b-c)(a+c-b)(b+c-a) \le 3abc \Leftrightarrow (3-2a)(3-2b)(3-2c) \le abc \Rightarrow3abc \ge 4(ab+bc+ac)-9[/tex] Thế vào (1) suy ra:
[tex]ab+bc+ac \le 3 (DONE!!!!!)[/tex]