4 replies to this topic

[trankienduc]

1. cho a,b,c >0 , a+b+c =3 . tìm GTNN $A=\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+c}$

2. cho a.b,c >0 . abc=1

cm $\frac{a^3}{a^2+ab+b^2}$ >= $\frac{2a-b}{3}$

Edited by trankienduc, 16-04-2016 - 19:29.

[lehakhiem212]

1.

Áp dụng cauchy - schwarz

A$\geq \frac{(a+b+c)^{2}}{2(a+b+c)}=\frac{3}{2}$

Dấu = khi a=b=c=1

Edited by lehakhiem212, 16-04-2016 - 19:34.

[ineX]

 

2. cho a.b,c >0 . abc=1

cm $\frac{a^3}{a^2+ab+b^2}$ >= $\frac{2a-b}{3}$

 

có phải đề bài là cho $abc>0$

Chứng mình: $\sum \frac{a^{3}}{a^{2}+b^{2}+ab}\geq \sum \frac{2a-b}{3}$

[ngocminhxd]

đặt a+b=x,b+c=y,c+a=z

=>a+b+c=$\frac{x+y+z}{2}=>a=\frac{y+z-x}{2}$

tt ta có $b=\frac{z+x-y}{2}$$

$c=\frac{y+x-z}{2}$$

ta có $(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})(a+b+c)\geq \frac{3}{2}(a+b+c)=> (\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b})(a+b+c)\geq \frac{9}{2} =>\frac{a^{2}+ab+ac}{b+c}+\frac{ab+b^{2}+bc}{a+c}+\frac{ac+bc+c^{2}}{a+b}\geq \frac{9}{2}=>\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}+a+b+c\geq \frac{9}{2}=>\frac{a^{2}}{b+c}+\frac{b^{2}}{c+a}+\frac{c^{2}}{a+b}\geq \frac{3}{2}$

[hieuhanghai]

2. cho a.b,c >0 . abc=1

cm $\frac{a^3}{a^2+ab+b^2}$ >= $\frac{2a-b}{3}$

Cái này biến đổi tương đương thui

Pt<=>$\frac{(a-b)^{2}}{3}.\frac{a+b}{a^{2}+ab+b^{2}}\geq 0 (luôn đúng do a,b \geq 0)$

Edited by hieuhanghai, 16-04-2016 - 20:04.