5 replies to this topic

[minhhien2001]

a,b,c$\epsilon \mathbb{R}$.a+b+c=3.c/m$a^2+b^2+c^2+ab+ac+bc\geqslant 6$

[rainbow99]

a,b,c$\epsilon \mathbb{R}$.a+b+c=3.c/m$a^2+b^2+c^2+ab+ac+bc\geqslant 6$

$a^{2}+b^{2}+c^{2}+ab+bc+ca=\frac{1}{2}\left [ (a+b)^{2}+(b+c)^{2}+(c+a)^{2} \right ]\geq \frac{1}{2}\left [ \frac{4(a+b+c)^{2}}{3} \right ]=6$

Edited by rainbow99, 02-04-2015 - 23:48.

[Nguyen Minh Hai]

a,b,c$\epsilon \mathbb{R}$.a+b+c=3.c/m$a^2+b^2+c^2+ab+ac+bc\geqslant 6$

Ta có:

$a^2+b^2+c^2+ab+bc+ca=\frac{1}{2}[(a+b)^2+(b+c)^2+(c+a)^2]\geq \frac{1}{6}(a+b+b+c+c+a)^2=\frac{2}{3}(a+b+c)^2=\frac{2}{3}.3^2=6$

[Duong Nhi]

 $\sum a^2 + \sum ab= (a+b+c)^2-(ab+ac+bc)\geq 3^2-\frac{3^2}{3}=6$

[Nguyen Minh Hai]

$a^{2}+b^{2}+c^{2}+ab+bc+ca=\frac{1}{2}\left [ (a+b)^{2}+(b+c)^{2}+(c+a)^{2} \right ]\geq \frac{1}{2}\left [ \frac{2(a+b+c)^{2}}{3} \right ]=6$

Chỗ màu đỏ là số $4$

[HoangVienDuy]

$(a+b+c)^{2}\geq 3(ab+bc+ca)\Rightarrow ab+bc+ca\leq 3$

$a^{2}+b^{2}+c^{2}+ab+bc+ca=(a+b+c)^{2}-ab-bc-ca\geq 9-3=6$

Edited by HoangVienDuy, 02-04-2015 - 21:53.