Tìm kiếm
Nam
Gửi bởi 2/ Cho a,b,c là các số nguyên dương. Chứng minh:
$\sum \frac{(b+c-a)^2}{(b+c)^2+a^2}\geq \frac{3}{5}$
Ta có: $A=\sum\frac{(b+c-a)^2}{(b+c)^2+a^2}$
<=> $A=\sum\frac{(b+c)^2+a^2-2a(b+c)}{(b+c)^2+a^2}$
<=> $A=\sum 1-\frac{2a(b+c)}{(b+c)^2+a^2}$
<=> $A=3 -\sum\frac{2a(b+c)}{(b+c)^2+a^2}$
Áp dụng bắt đẳng thức $AM-GM$, ta có:
$a^2+\frac{(b+c)^2}{4}\geq a(b+c)$
<=> $a^2+(b+c)^2\geq a(b+c)+\frac{3}{4}(b+c)^2$
Do đó: $A\geq 3-\sum\frac{2a(b+c)}{a(b+c)+\frac{3}{4}.(b+c)^2}$
<=> $A\geq 3-\sum 2-\frac{\frac{3}{2}.(b+c)^2}{a(b+c)+\frac{3}{4}.(b+c)^2}$
<=> $A\geq\sum\frac{\frac{3}{2}(b+c)^2}{a(b+c)+\frac{3}{4}.(b+c)^2}-3$
<=> $A\geq\frac{3}{2}.B-3 (1)$
Ta có: $B=\sum\frac{(b+c)^2}{(b+c)(a+\frac{3}{4}(b+c))}$
<=>$B=\sum\frac{b+c}{a+\frac{3}{4}(b+c)}$
<=>$\frac{1}{4}B+3=\sum{\frac{\frac{1}{4}(b+c)}{a+\frac{3}{4}(b+c)}+1}$
<=>$\frac{1}{4}B+3=(a+b+c).\sum\frac{1}{a+\frac{3}{4}.(b+c)}\geq (a+b+c).\frac{9}{\sum{a+\frac{3}{4}(b+c)}}$
Lại có: $VP=(a+b+c).\frac{9}{\frac{5}{2}(a+b+c)}=\frac{18}{5}$.
Do đó, $\frac{1}{4}B+3\geq\frac{18}{5}$
<=> $B\geq(\frac{18}{5}-3).4=\frac{12}{5} (2)$
Từ $(1),(2)$, ta có: $A\geq \frac{3}{2}.\frac{12}{5}-3=\frac{3}{5}$
=> $Q.E.D$
Vậy $\sum \frac{(b+c-a)^2}{(b+c)^2+a^2}\geq \frac{3}{5}$
Dấu $"="$ xảy ra <=> $a=b=c\in\mathbb{Z_+}$
- minhduc2000, Ngoc Hung và Namthemaster1234 thích
