Cho $a , b , c , d > 0$. Chứng minh rằng :
$\sum \frac{a}{b + c + d} \geq \frac{4}{3}$.
$\sum \frac{a}{b + c + d} \geq \frac{4}{3}$
Started By tramyvodoi, 12-02-2013 - 20:20
#1
Posted 12-02-2013 - 20:20
- Khanh 6c Hoang Liet and nguyen tien dung 98 like this
#2
Posted 12-02-2013 - 20:31
Cho $a , b , c , d > 0$. Chứng minh rằng :
$\sum \frac{a}{b + c + d} \geq \frac{4}{3}$.
$\sum \frac{a}{b+c+d}=\sum \frac{a^{2}}{ab+ac+ad}\geq \frac{(a+b+c+d)^{2}}{2ab+2ac+2ad+2bc+2bd+2cd}$
Do đó ta chỉ cần chứng minh
$(a+b+c+d)^{2}\geq \frac{8}{3}(ab+bc+cd+da+ac+bd)$
Chứng minhTa có
$3(a+b+c+d)^{2}\geq 8(ab+bc+cd+da+ac+bd)$
$\Leftrightarrow 3a^{2}+3b^{2}+3c^{2}+3d^{2}\geq 2(ab+bc+cd+da+ac+bd)$
$\Leftrightarrow (a-b)^{2}+(b-c)^{2}+(c-d)^{2}+(d-a)^{2}+(a-c)^{2}+(b-d)^{2}\geq 0$
- Tienanh tx, Primary and nguyen tien dung 98 like this
#3
Posted 12-02-2013 - 20:46
Cách khác :
Đặt $x = b + c + d$ $,$ $y = c + d + a$ $,$ $z = d + a + b$ $,$ $t = a + b + c$.
Do đó :
$a + b + c + d = \frac{x + y + z + t}{3}$.
Nên :
$a = \frac{-2x + y + z + t}{3}$ $,$
$b = \frac{x - 2y + z + t}{3}$ $,$
$c = \frac{x + y - 2z + t}{3}$ $,$
$d = \frac{x + y + z - 2t}{3}$.
Ta có :
$\sum \frac{a}{b + c + d} \geq \frac{-2x + y + z + t}{3} + \frac{x - 2y + z + t}{3} + \frac{x + y - 2z + t}{3} + \frac{x + y + z - 2t}{3} = -\frac{8}{3} + \frac{1}{3}\left [ \sum \left ( \frac{x}{y} + \frac{y}{x} \right ) + 12 \right ] \geq -\frac{8}{3} + \frac{1}{3}\left [ \sum \frac{\left ( x - y \right )^{2}}{xy} + 12 \right ] \geq -\frac{8}{3} + \frac{1}{3}.12 = \frac{4}{3}$.
Ta có $\text{đpcm}$.
Đặt $x = b + c + d$ $,$ $y = c + d + a$ $,$ $z = d + a + b$ $,$ $t = a + b + c$.
Do đó :
$a + b + c + d = \frac{x + y + z + t}{3}$.
Nên :
$a = \frac{-2x + y + z + t}{3}$ $,$
$b = \frac{x - 2y + z + t}{3}$ $,$
$c = \frac{x + y - 2z + t}{3}$ $,$
$d = \frac{x + y + z - 2t}{3}$.
Ta có :
$\sum \frac{a}{b + c + d} \geq \frac{-2x + y + z + t}{3} + \frac{x - 2y + z + t}{3} + \frac{x + y - 2z + t}{3} + \frac{x + y + z - 2t}{3} = -\frac{8}{3} + \frac{1}{3}\left [ \sum \left ( \frac{x}{y} + \frac{y}{x} \right ) + 12 \right ] \geq -\frac{8}{3} + \frac{1}{3}\left [ \sum \frac{\left ( x - y \right )^{2}}{xy} + 12 \right ] \geq -\frac{8}{3} + \frac{1}{3}.12 = \frac{4}{3}$.
Ta có $\text{đpcm}$.
- banhgaongonngon, Tienanh tx, Khanh 6c Hoang Liet and 2 others like this
#4
Posted 12-02-2013 - 21:15
cách khác(hi vọng ko trùng):Cho $a , b , c , d > 0$. Chứng minh rằng :
$\sum \frac{a}{b + c + d} \geq \frac{4}{3}$.
$\sum \frac{a}{b+c+d}\geq \frac{4}{3}\Leftrightarrow (\sum a)(\sum \frac{1}{b+c+d})\geq \frac{16}{3}\Leftrightarrow \frac{1}{3}(\sum b+c+d)(\sum \frac{1}{b+c+d})\geq \frac{16}{3}$(luôn đúng)
$\text{Cứ làm việc chăm chỉ trong im lặng}$
$\text{Hãy để thành công trở thành tiếng nói của bạn}$
#5
Posted 09-03-2013 - 20:44
Ta có:$3(\sum \frac{a}{b+c+d}+4)=3(\sum \frac{a+b+c+d}{b+c+d})=3\(a+b+c+d)\sum\frac{1}{b+c+d}=3(a+b+c+d)\sum \frac{1}{b+c+d}\geq 16 \Rightarrow \sum \frac{a}{b+c+d}\leq \frac{4}{3}$
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