B
bigbang195
Cho a,b,c>0 thõa abc=1 CMR
[TEX]RHS=\frac{4\sum [a(a+c)(a+b)]+3(a+b)(b+c)(a+c)}{(a+b)(b+c)(a+c)}[/TEX]
[TEX]=\frac{4\sum[a^3+a^2b+b^2a+abc]+3\prod (a+b)}{\prod (a+b)}[/TEX]
[TEX]=\frac{4\sum a^3+4\prod (a+b)+4+3\prod (a+b)}{\prod (a+b)}[/tex]
[TEX]=\frac{4\sum a^3 +12\prod (a+b)+4-5\prod (a+b)}{\prod (a+b)}[/TEX]
cần CM
[TEX](a+b+c)^2\prod (a+b) \ge 4\sum a^3 +12\prod (a+b)+4-5\prod (a+b)[/TEX]
[TEX]=4(a+b+c)^3-5\prod (a+b)+4[/TEX]
tức là
[TEX]p^2(qp-1)\ge 4p^3-5(qp-1)+4=4p^3-5qp+9[/TEX]
hay[TEX] p^3q-p^2-4p^3+5qp-9 \ge 0[/TEX]
ta có[TEX] p \ge 3 , q \ge 3[/TEX]
nếu [TEX]q \ge p[/TEX] thì
[TEX]p^3q-p^2-4p^3+5qp-9 \ge p^4-4p^3+5p^2-p^2-9=p^4-4p^3+4p^2-9=[/TEX]
[TEX](p-3)(p+1)(p^2-2p+3) \ge 0[/TEX]
Nếu [TEX]p \ge q[/TEX] thì .
Quay lại BDT ban theo Cauchy-Schwarz thì
[TEX]\frac{1}{x}+\frac{1}{y} \ge \frac{4}{x+y}[/TEX]
[TEX]VP \le \sum \frac{a+b}{c}+3 =\sum ab(a+b)+3=pq-3+3=pq[/TEX]
nên [TEX]p^2 \ge pq[/TEX]đúng!
Chứng Minh hoàn tất
Chứ thích
[TEX]p=a+b+c,q=ab+bc+ac[/TEX]
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