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 $\Leftrightarrow$ $a=b=c$

Bạn cần đăng nhập để có thể gửi đáp án

Chat chit và chém gió
  • Việt EL: ... 8/21/2017 8:20:01 AM
  • Việt EL: ... 8/21/2017 8:20:03 AM
  • wolf linhvân: 222 9/17/2017 7:22:51 AM
  • dominhdai2k2: u 9/21/2017 7:31:33 AM
  • arima sama: helllo m 10/8/2017 6:49:28 AM
  • ๖ۣۜGemღ: Mọi người có thắc mắc hay cần hỗ trợ gì thì gửi tại đây nhé https://goo.gl/dCdkAc 12/6/2017 8:53:25 PM
  • anhkind: hi mọi người mk là thành viên mới nè 12/28/2017 10:46:02 AM
  • anhkind: party 12/28/2017 10:46:28 AM
  • Rushia: . 2/27/2018 2:09:24 PM
  • Rushia: . 2/27/2018 2:09:25 PM
  • Rushia: . 2/27/2018 2:09:25 PM
  • Rushia: . 2/27/2018 2:09:26 PM
  • Rushia: . 2/27/2018 2:09:26 PM
  • Rushia: . 2/27/2018 2:09:26 PM
  • Rushia: . 2/27/2018 2:09:26 PM
  • Rushia: . 2/27/2018 2:09:27 PM
  • Rushia: . 2/27/2018 2:09:27 PM
  • Rushia: . 2/27/2018 2:09:28 PM
  • Rushia: . 2/27/2018 2:09:28 PM
  • Rushia: . 2/27/2018 2:09:28 PM
  • Rushia: . 2/27/2018 2:09:29 PM
  • Rushia: . 2/27/2018 2:09:29 PM
  • Rushia: . 2/27/2018 2:09:29 PM
  • Rushia: . 2/27/2018 2:09:29 PM
  • Rushia: . 2/27/2018 2:09:30 PM
  • Rushia: . 2/27/2018 2:09:30 PM
  • Rushia: . 2/27/2018 2:09:31 PM
  • Rushia: .. 2/27/2018 2:09:31 PM
  • Rushia: . 2/27/2018 2:09:32 PM
  • Rushia: . 2/27/2018 2:09:32 PM
  • Rushia: . 2/27/2018 2:09:32 PM
  • Rushia: . 2/27/2018 2:09:32 PM
  • Rushia: . 2/27/2018 2:09:33 PM
  • Rushia: . 2/27/2018 2:09:33 PM
  • Rushia: . 2/27/2018 2:09:33 PM
  • Rushia: . 2/27/2018 2:09:34 PM
  • ๖ۣۜBossღ: c 3/2/2018 9:20:18 PM
  • nguoidensau2k2: hello 4/21/2018 7:46:14 PM
  • ☼SunShine❤️: Vẫn vậy <3 7/31/2018 8:38:39 AM
  • ☼SunShine❤️: Bên này text chữ vẫn đẹp nhất <3 7/31/2018 8:38:52 AM
  • ☼SunShine❤️: @@ lại càng đẹp <3 7/31/2018 8:38:59 AM
  • ☼SunShine❤️: Hạnh phúc thế sad mấy câu hỏi vớ vẩn hồi trẩu vẫn hơn 1k xem 7/31/2018 8:41:00 AM
  • tuyencr123: vdfvvd 3/6/2019 9:30:53 PM
  • tuyencr123: dv 3/6/2019 9:30:53 PM
  • tuyencr123: d 3/6/2019 9:30:54 PM
  • tuyencr123: dv 3/6/2019 9:30:54 PM
  • tuyencr123: d 3/6/2019 9:30:54 PM
  • tuyencr123: d 3/6/2019 9:30:55 PM
  • tuyencr123: đ 3/6/2019 9:30:55 PM
  • tuyencr123: đ 3/6/2019 9:30:56 PM
  • tuyencr123: d 3/6/2019 9:30:56 PM
  • tuyencr123: d 3/6/2019 9:30:56 PM
  • tuyencr123: d 3/6/2019 9:30:56 PM
  • tuyencr123: d 3/6/2019 9:30:56 PM
  • tuyencr123: d 3/6/2019 9:30:56 PM
  • tuyencr123: d 3/6/2019 9:30:57 PM
  • tuyencr123: d 3/6/2019 9:30:57 PM
  • tuyencr123: d 3/6/2019 9:30:57 PM
  • tuyencr123: d 3/6/2019 9:30:57 PM
  • tuyencr123: d 3/6/2019 9:30:57 PM
  • tuyencr123: d 3/6/2019 9:30:58 PM
  • tuyencr123: đ 3/6/2019 9:30:58 PM
  • tuyencr123: d 3/6/2019 9:30:58 PM
  • tuyencr123: d 3/6/2019 9:30:58 PM
  • tuyencr123: d 3/6/2019 9:30:59 PM
  • tuyencr123: d 3/6/2019 9:30:59 PM
  • tuyencr123: d 3/6/2019 9:30:59 PM
  • tuyencr123: d 3/6/2019 9:30:59 PM
  • tuyencr123: d 3/6/2019 9:30:59 PM
  • tuyencr123: d 3/6/2019 9:31:00 PM
  • tuyencr123: d 3/6/2019 9:31:00 PM
  • tuyencr123: d 3/6/2019 9:31:00 PM
  • tuyencr123: d 3/6/2019 9:31:00 PM
  • tuyencr123: đ 3/6/2019 9:31:01 PM
  • tuyencr123: d 3/6/2019 9:31:01 PM
  • tuyencr123: đ 3/6/2019 9:31:01 PM
  • tuyencr123: d 3/6/2019 9:31:02 PM
  • tuyencr123: d 3/6/2019 9:31:02 PM
  • tuyencr123: d 3/6/2019 9:31:02 PM
  • tuyencr123: d 3/6/2019 9:31:02 PM
  • tuyencr123: d 3/6/2019 9:31:02 PM
  • tuyencr123: d 3/6/2019 9:31:03 PM
  • tuyencr123: d 3/6/2019 9:31:03 PM
  • tuyencr123: d 3/6/2019 9:31:03 PM
  • tuyencr123: d 3/6/2019 9:31:03 PM
  • tuyencr123: d 3/6/2019 9:31:04 PM
  • tuyencr123: d 3/6/2019 9:31:04 PM
  • tuyencr123: d 3/6/2019 9:31:04 PM
  • tuyencr123: d 3/6/2019 9:31:04 PM
  • tuyencr123: d 3/6/2019 9:31:05 PM
  • tuyencr123: đ 3/6/2019 9:31:05 PM
  • tuyencr123: bb 3/6/2019 9:31:06 PM
  • tuyencr123: b 3/6/2019 9:31:06 PM
  • tuyencr123: b 3/6/2019 9:31:06 PM
  • tuyencr123: b 3/6/2019 9:31:07 PM
  • tuyencr123: b 3/6/2019 9:31:38 PM
  • Tríp Bô Hắc: cho hỏi lúc đăng câu hỏi em có thấy dòng cuối là tabs vậy ghi gì vào tabs vậy ạ 7/15/2019 7:36:37 PM
  • khanhhuyen2492006: hi 3/19/2020 7:33:03 PM
  • ngoduchien36: hdbnwsbdniqwjagvb 11/17/2020 2:36:40 PM
  • tongthiminhhangbg: hello 6/13/2021 2:22:13 PM
Đăng nhập để chém gió cùng mọi người
  • hoàng anh thọ
  • Thu Hằng
  • Xusint
  • HọcTạiNhà
  • lilluv6969
  • ductoan933
  • Tiến Thực
  • my96thaibinh
  • 01668256114abc
  • Love_Chishikitori
  • meocon_loveky
  • gaprodianguc95
  • smallhouse253
  • hangnguyen.hn95.hn
  • nguyencongtrung9744
  • tart
  • kto138
  • dphonglkbq
  • ๖ۣۜPXM๖ۣۜMinh4212♓
  • huyhieu10.11.1999
  • phungduyen1403
  • lalinky.ltml1212
  • trananhvan12315
  • linh31485
  • thananh133
  • Confusion
  • Hàn Thiên Dii
  • •♥•.¸¸.•♥•Furin•♥•.¸¸.•♥•
  • dinhtuyetanh000
  • LeQuynh
  • tuanmotrach
  • bac1024578
  • truonglinhyentrung
  • Lê Giang
  • Levanbin147896325
  • anhquynhthivu
  • thuphuong30012003