Bất Động (ACAMOPHOMADADY 2016-2017)

Question

Cho

$a,b,c>0$ thỏa mãn:

$abc=1$ . Chứng minh rằng:

$\sum \frac{a^2(b+1)}{b(a^2+ab+b^2)}\ge \frac{6}{a+b+c}$

★·.·´¯`·.·★Poseidon★·.·´¯`·.·★ · Answer 1 · 8/19/2016 8:51:14 PM

Xét

$\frac{1}{b}-\frac{a^2.(b+1)}{b(a^2+ab+b^2)}=\frac{a+b-a^2}{a^2+ab+b^2}$

$\Rightarrow \frac{a^2.(b+1)}{b(a^2+ab+b^2)}=\frac{1}{b}-\frac{a+b-a^2}{a^2+ab+b^2}=\frac{1}{b}-\frac{a+b}{a^2+ab+b^2}+\frac{a^2}{a^2+ab+b^2}$

Ta có:

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

+) Đặt

$M=\sum \frac{a+b}{a^2+b^2+ab}=\sum \frac{a+b}{\frac{3}{4}.(a+b)^2+\frac{1}{4}.(a-b)^2} \leq \sum\frac{a+b}{\frac{3}{4}.(a+b)^2}$

$\Rightarrow M \leq \frac{4}{3}.\sum \frac{1}{a+b} \leq \frac{4}{3}.\sum \frac{1}{4}.(\frac{1}{a}+\frac{1}{b})$

$\Rightarrow M\leq \frac{2}{3}.(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$

+) Đặt

$N=\sum \frac{a^2}{a^2+ab+b^2}$

Áp dụng

$Cauchy-Schwart$ ta có :

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

$\Rightarrow N\geq \frac{(a+b+c)^2}{2.(a+b+c)^2-9}$ (Do

$ab+bc+ca\geq 3\sqrt{a^2+b^2+c^2}=3$ )

Do đó :

$VT\geq \sum \frac{1}{a}-\frac{2}{3}.\sum \frac{1}{a}+\frac{(a+b+c)^2}{2.(a+b+c)^2-9}$

$VT\geq \frac{1}{3}.\sum \frac{1}{a}+\frac{(a+b+c)^2}{2(a+b+c)^2-9}\geq \frac{3}{a+b+c}+\frac{(a+b+c)^2}{2(a+b+c)^2-9}=A$

Ta sẽ chứng minh

$A\geq \frac{6}{a+b+c} \Leftrightarrow \frac{(a+b+c)^2}{2.(a+b+c)^2-9} \geq \frac{3}{(a+b+c)}$

Đặt

$a+b+c=t$ . Cần chứng minh

$\frac{t^2}{2t^2-9} \geq \frac{3}{t}\Leftrightarrow (t-3)(t^2-3t+9)\geq 0$

Luôn đúng do

$t=a+b+c\geq 3\sqrt[3]{abc}=3$

Vậy bất đẳng thức được chứng minh. Dấu bằng khi

$a=b=c=1$

1 Đáp án