2 replies to this topic

[Oral1020]

$\boxed{1}$Cho $a,b,c >0$ thõa mản $a+b+c=1$.Chứng minh rằng:

a)$2(a^3+b^3+c^3)+3abc \ge a^2+b^2+c^2$

b)$\dfrac{a^3+b^3+c}{1-c}+\dfrac{b^3+c^3+a}{1-a}+\dfrac{c^3+a^3+b}{1-b} \ge \dfrac{a^2+b^2+1}{2}+\dfrac{a^2+c^2+1}{2}+\dfrac{c^2+b^2+1}{2}$

$\boxed{2}$Cho $a,b,c>0$ thỏa mãn $a^4+b^4+c^4 \ge \dfrac{t^4}{27}$.Tìm GTNN của $A$ theo $t$
$A=\dfrac{a^2}{b+c}+\dfrac{b^2}{a+c}+\dfrac{c^2}{a+b}$

[hoangvtvpvn]

ĐPCM$\Leftrightarrow 2\sum a^{3}+3abc\geq \sum a^{2}\sum a \Leftrightarrow \sum a^{3}+3abc\geq \sum a^{2}b$(luôn đúng theo schur)

Do đó ta có đpcm

[DarkBlood]

$\boxed{1}$Cho $a,b,c >0$ thõa mản $a+b+c=1$.Chứng minh rằng:

b)$\dfrac{a^3+b^3+c}{1-c}+\dfrac{b^3+c^3+a}{1-a}+\dfrac{c^3+a^3+b}{1-b} \ge \dfrac{a^2+b^2+1}{2}+\dfrac{a^2+c^2+1}{2}+\dfrac{c^2+b^2+1}{2}$

Ta có:

$VT=\sum \frac{a^3+b^3+c}{1-c}=\sum \frac{(a+b)(a^2-ab+b^2)+c}{a+b}=$

$=2\sum a^2-\sum ab+\sum \frac{c}{a+b}\geq \sum a^2+\sum \frac{3}{2}=VP$

Dấu bằng xảy ra khi $\left\{\begin{matrix} a=b=c\\ a+b+c=1 \end{matrix}\right.\ \ \Longleftrightarrow\ \ a=b=c=\frac{1}{3}$

Edited by Hoang Huy Thong, 11-04-2013 - 18:17.