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[25 minutes]

Ch0 các số thực $a,b,c$ thỏa mãn $a^2+b^2+c^2=3$

Chứng minh rằng $a+b+c-abc\geqslant -\sqrt{6}$

 

[daicahuyvn]

$P=a+b+c-abc$

 $|a|\ge |b| \ge| c | ;|abc|\le 1\Rightarrow |bc|\le 1\Rightarrow -1\le bc \le 1 $

$P^2=[(b+c)+a(1-bc)]^2\le [a^2+(b+c)^2][(1-bc)^2+1]$

$bc=t(-1\le t\le 1).P^2\le (3+2t)(t^2-2t+2)=2t^3-t^2-2t+6=f(t)$

$f'(t)=2(3t^2-t-1). f(t)\le f(\frac{1-\sqrt{13}}{6})=\frac{305+13\sqrt{13}}{54}\approx 6,516151233$

$P\ge - \sqrt{\frac{305+13\sqrt{13}}{54}}$

$a=b=-\sqrt{\frac{5+\sqrt{13}}{6}},c=\sqrt{\frac{4-\sqrt{13}}{3}} \rightarrow P=- \sqrt{\frac{305+13\sqrt{13}}{54}}$

$\min P=- \sqrt{\frac{305+13\sqrt{13}}{54}}$

Edited by daicahuyvn, 13-08-2013 - 13:52.