Chủ đề này có 4 trả lời

[tramyvodoi]

Cho $a , b , c , d > 0$. Chứng minh rằng :
$\sum \frac{a}{b + c + d} \geq \frac{4}{3}$.

[banhgaongonngon]

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 minh
Ta có

$3(a+b+c+d)^{2}\geq 8(ab+bc+cd+da+ac+bd)$

$\Leftrightarrow 3a^{2}+3b^{2}+3c^{2}+3d^{2}\geq 2(ab+bc+cd+da+ac+bd)$

$\Leftrightarrow (a-b)^{2}+(b-c)^{2}+(c-d)^{2}+(d-a)^{2}+(a-c)^{2}+(b-d)^{2}\geq 0$

[tramyvodoi]

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 $\text{đpcm}$.

[Mrnhan]

Cho $a , b , c , d > 0$. Chứng minh rằng :
$\sum \frac{a}{b + c + d} \geq \frac{4}{3}$.

cách khác(hi vọng ko trùng):
$\sum \frac{a}{b+c+d}\geq \frac{4}{3}\Leftrightarrow (\sum a)(\sum \frac{1}{b+c+d})\geq \frac{16}{3}\Leftrightarrow \frac{1}{3}(\sum b+c+d)(\sum \frac{1}{b+c+d})\geq \frac{16}{3}$(luôn đúng)

[buiminhhieu]

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}$