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[Tham Lang]

Bài toán :
Cho $a+b+c+d+e+f=1, ace+bdf \ge \dfrac{1}{108}; a,b,c,d,e,f >0$
Chứng minh rằng :
$$abc+bcd+cde+def+efa+fab \le \dfrac{1}{36}$$

[tim1nuathatlac]

Bài toán :
Cho $a+b+c+d+e+f=1, ace+bdf \ge \dfrac{1}{108}; a,b,c,d,e,f >0$
Chứng minh rằng :
$$abc+bcd+cde+def+efa+fab \le \dfrac{1}{36}$$

ta có $ace+bdf+abc+bcd+cde+def+efa+fab=\left ( a+d \right )\left ( b+e \right )\left ( c+f \right )\leq \frac{\left ( a+b+c+d+e+f \right )^{3}}{27}=\frac{1}{27}$

$\Rightarrow bcd+abc+cde+def+efa+fba\leq \frac{1}{27}-\frac{1}{108}=\frac{1}{36}$
dấu = xảy ra khi $a=b=c=d=e=f=\frac{1}{6}$
( Phạm Kim Hùng )

Edited by tim1nuathatlac, 01-09-2012 - 09:32.