Cho a,b,c là các số thực dương. 
CMR:$\frac{a}{\sqrt{b+c}}+\frac{b}{\sqrt{a+c}}+\frac{c}{\sqrt{a+b}}\geqslant \frac{1}{\sqrt{2}}(\sqrt{a}+\sqrt{b}+\sqrt{c})$
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 (*)&(**)$\Rightarrow đpcm$
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
Áp dụng bdt holder :
$VT^2.(a(b+c)+b(c+a)+c(a+b)) \ge (a+b+c)^3$
$\leadsto VT^2 \ge \frac{(a+b+c)^3}{2(ab+bc+ca)} \ge \frac{3(a+b+c)}{2} \ge \frac{(\sqrt{a}+\sqrt{b}+\sqrt{c})^2}{2}$
$\leadsto dpcm$

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
E có thể tham khảo :) http://www.slideshare.net/Truonghocso/19-phng-phap-chng-minh-bt-ng-thc
thích thê đấy ==" –  ๖ۣۜDemonღ 04-07-16 05:08 PM
cái này cho vào phần bình luận em ơi –  ๖ۣۜDevilღ 04-07-16 04:03 PM
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 $\Leftrightarrow$ $a=b=c$

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