Cho e 3 cách lm ạ ..... Thaks nhiều

Question

Cho a,b,c là các số thực dương.

CMR:

$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}}+\frac{c}{\sqrt{a+b}}\geqslant \frac{1}{\sqrt{2}}(\sqrt{a}+\sqrt{b}+\sqrt{c})$

Bloody's Rose · Answer 1 · 7/4/2016 9:10:51 PM

C1:Ta CM:

$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}}+\frac{c}{\sqrt{a+b}}\geq\sqrt{\frac{3}{2}(a+b+c)}\geq\frac{1}{\sqrt{2}}(\sqrt{a}+\sqrt{b}+\sqrt{c})$

+)

$\Sigma \frac{a}{\sqrt{b+c}}\geq\sqrt{\frac{3}{2}(a+b+c)}$ (1)

Do tính thuần nhất của BĐT,ta chuẩn hóa:

$a+b+c=\frac{3}{2}$

Ta cần CM:

$\Sigma \frac{a}{\sqrt{b+c}}\geq\frac{3}{2}$

ÁD BĐT C-S:

$VT\geq\frac{(a+b+c)^{2}}{\Sigma a\sqrt{b+c}}=\frac{9}{4(\Sigma a\sqrt{b+c})}$

Ta chỉ cần CM:

$\Sigma a\sqrt{b+c}\leq \frac{3}{2}$

Thật vậy:ÁD BĐT C-S:

$a\sqrt{b+c}+b\sqrt{c+a}+c\sqrt{a+b}\leq\sqrt{(a+b+c)\left[ {a(b+c)+b(c+a)+c(a+b)} \right]}$

=

$\sqrt{3(ab+bc+ca)}\leq a+b+c=\frac{3}{2}$

$\Rightarrow$ (1) đúng(*)

+)ÁD BĐT Bunhiacopxki::

$\sqrt{3(a+b+c)}\geq\Sigma \sqrt{a}$

$\Rightarrow$ (2) đúng(**)

Từ (*)&(**)

$\Rightarrow đpcm$

score 16 · Answer 2

Ta có

$\frac{\sqrt a+\sqrt b+\sqrt c}{\sqrt 2} \overset{C-S}\le \frac{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}{2}$

Nên chỉ cần cm

$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \ge \frac{\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}}{2}$

$\leadsto \sum\left(\frac{a}{\sqrt{b+c}}-\frac{\sqrt{b+c}}{2}\right) \ge0$

$\leadsto \sum\frac{(a-b)+(a-c)}{2\sqrt{b+c}} \ge0$

$\leadsto (a-b)\left( \tfrac{1}{\sqrt{b+c}}-\tfrac{1}{\sqrt{c+a}}\right)+(b-c)\left(\tfrac{1}{\sqrt{c+a}}-\tfrac{1}{\sqrt{a+b}} \right)+(c-a)\left(\tfrac{1}{\sqrt{a+b}}-\tfrac{1}{\sqrt{b+c}} \right) \ge0$

KMTTQ,giả sử

$a\ge b \ge c$ , dễ dàng cm

$\; (c-a)\left(\tfrac{1}{\sqrt{a+b}}-\tfrac{1}{\sqrt{b+c}} \right) \ge0$

Khi đó chỉ cần cm

$(a-b)\left( \tfrac{1}{\sqrt{b+c}}-\tfrac{1}{\sqrt{c+a}}\right)+(b-c)\left(\tfrac{1}{\sqrt{c+a}}-\tfrac{1}{\sqrt{a+b}} \right) \ge0$

$\leftrightsquigarrow \frac{(a-b)(\sqrt{c+a}-\sqrt{b+c})}{\sqrt{(b+c)(c+a)}}+\frac{(b-c)(\sqrt{a+b}-\sqrt{a+c})}{\sqrt{(c+a)(a+b)}} \ge0$

$\leftrightsquigarrow\frac{(a-b)^2}{(\sqrt{c+a}+\sqrt{b+c})\sqrt{b+c}}+\frac{(b-c)(a-c)}{(\sqrt{a+b}+\sqrt{b+c})\sqrt{a+b}} \ge0$

BDT cuối luôn đúng suy ra dpcm

score 15 · Answer 3

Bình chọn tăng 15 Bình chọn giảm

Áp dụng bdt holder :

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

$\leadsto VT^2 \ge \frac{(a+b+c)^3}{2(ab+bc+ca)} \ge \frac{3(a+b+c)}{2} \ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{2}$

$\leadsto dpcm$

score 0 · Answer 4

Không mất tính tổng quát, ta chuẩn hóa

$a+b+c=3$

Xét

$\frac{a}{\sqrt{b+c}} + \frac{b}{\sqrt{c+a}}+\frac{c}{\sqrt{a+b}} \geq \frac{1}{\sqrt{2}}(\sqrt{a}+\sqrt{b}+\sqrt{c})$

$\Leftrightarrow$

$\frac{\sqrt{2}a}{\sqrt{b+c}} + \frac{\sqrt{2}b}{\sqrt{c+a}} + \frac{\sqrt{2}c}{\sqrt{a+b}} \geq \sqrt{a} + \sqrt{b} + \sqrt{c}$

$\Leftrightarrow$

$\frac{2a}{\sqrt{2(b+c)}} + \frac{2b}{\sqrt{2(c+a)}} + \frac{2c}{\sqrt{2(a+b)}} \geq \sqrt{a} + \sqrt{b} + \sqrt{c}$

Xét:

$A=\frac{2a}{\sqrt{2(b+c)}} + \frac{2b}{\sqrt{2(c+a)}} + \frac{2c}{\sqrt{a+b}}$ :

$\frac{A}{4}=\frac{a}{2\sqrt{2(b+c})}+\frac{b}{2\sqrt{2(c+a)}}+\frac{c}{2\sqrt{2(a+b)}}$

$\geq$

$\frac{a}{b+c+2} + \frac{b}{c+a+2} + \frac{c}{a+b+2}$

$\frac{A}{4} + 3 \geq (a+b+c+2)[\frac{1}{b+c+2}+\frac{1}{c+a+2}+\frac{1}{a+b+2}]$

Áp dụng bất đẳng thức Bunhia-Copxki ta có:

$\frac{1}{b+c+2} + \frac{1}{c+a+2} + \frac{1}{a+b+2} \geq \frac{(1+1+1)^2}{2a+2b+2c+2+2+2}=\frac{3}{4}$

Nên,

$\frac{A}{4} \geq \frac{5.3}{4} - 3 = \frac{3}{4}$

$\Leftrightarrow A \geq 3$ (1)

Áp dụng bất đẳng thức Bunhia-Copxki:

$(a+b+c)(1+1+1)=9 \geq (\sqrt{a} + \sqrt{b} + \sqrt{c})^2$

$\Leftrightarrow$

$\sqrt{a} + \sqrt{b} + \sqrt{c} \leq 3$ (2)

Từ (1) và (2) có thể suy ra bất đẳng thức luôn đúng

Dấu bằng xảy ra

$\Leftrightarrow$

$a=b=c$

❄⊰๖ۣۜNgốc๖ۣۜ ⊱ ❄ · Answer 5 · 7/5/2016 9:12:13 AM

Ta có:

$A=\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}}+\frac{c}{\sqrt{a+b}}$

$=(a+b+c)(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{a+c}}+\frac{1}{\sqrt{a+b}})-(\sqrt{b+c}+\sqrt{a+c}+\sqrt{a+b})$

Theo bất đẳng thức Cauchy-Schwarz ta có :

$(a+b+c)(\frac{1}{\sqrt{b+c}}+\frac{1}{\sqrt{a+c}}+\frac{1}{\sqrt{a+b}})\geqslant \frac{9(a+b+c)}{\sqrt{b+c}+\sqrt{a+c}+\sqrt{a+b}}$

Theo bất đẳng thức AM-GM ta có:

$\sqrt{b+c}+\sqrt{a+c}+\sqrt{a+b}\leq \sqrt{3.2.(a+b+c)}$

$\Rightarrow A\geq \frac{9(a+b+c)}{\sqrt{3.2.(a+b+c)}}-\sqrt{3.2.(a+b+c)}=\frac{\sqrt{3(a+b+c)}}{\sqrt{2}}\geq \frac{1}{\sqrt{2}}(\sqrt{a}+\sqrt{b}+\sqrt{c})$ (đpcm)

Đẳng thức xảy ra khi và chỉ khi a = b = c

Em lm đc 1 cách này thui .........haizzzz