$\dpi{150} \small \:hinh \:nhu \:la \:nam \:1766 \:dung \:ko \: a\: \: \:$
- bangbang1412 và babystudymaths thích
Gửi bởi ngoctruong236 trong 27-07-2013 - 18:11
$\dpi{150} \small \:hinh \:nhu \:la \:nam \:1766 \:dung \:ko \: a\: \: \:$
Gửi bởi ngoctruong236 trong 27-07-2013 - 18:06
Gửi bởi ngoctruong236 trong 27-07-2013 - 18:05
$\dpi{150} \small Bai nay thuc chat la bai toan co ban sau (a+b+c)^3\geq \frac{27}{4}(ab^2+bc^2+ca^2+abc)$
Gửi bởi ngoctruong236 trong 27-07-2013 - 18:01
$\dpi{150} \small \:Theo \:Bdt \: Holder\: ta\: co\: \sum \sqrt{\frac{b+c}{a}})^2\left [ \sum \frac{1}{a^2(b+c)} \right ]\geq(\sum \frac{1}{}a)^3 \:Do \:đo \:ta \:chi \:can \:CM \ \sum \frac{1}{a})^3\geq \frac{6(a+b+c)}{\sqrt[3]{abc}}\sum \frac{1}{a^2(b+c)}.\:Đặt \:a=\frac{1}{x} ,b= \frac{1}{y},c= \frac{1}{z}\:Bdt\Leftrightarrow \:(x+y+z)^3\geq \6(xy+yz+zx)(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y})\:hay \:\frac{(x+y+z)^3}{\sqrt[3]{xyz}} \geq 6(x^2+y^2+z^2)+6xyz(\frac{1}{y+z}+\frac{1}{x+z}+\frac{1}{x+y})\:Áp \dụng \:Bdt Cauchy-Schwarz \: ,\:ta \:có \:6xyz(\frac{1}{y+z}+\frac{1}{z+x}+\frac{1}{x+y})\leq 6xyz(\frac{1}{4x}+\frac{1}{4y}+\frac{1}{4y}+\frac{1}{4z}+\frac{1}{4z}+\frac{1}{4x})= 3(xy+yz+zx)\:\frac{(x+y+z)^4}{\sqrt[3]{xyz}}\geq \3(x+y+z)^3. \:Như \:vậy \:ta \:chỉ \:cần \:CM \::3(x+y+z)^3\geq 6(x^2+y^2+z^2)+3(xy+yz+zx).Sau khi thu gọn ta dc dpcm.Đay \:là \:cách \: làm\:của \:em \moi : \:nguoi\:xem \:ho \:em \:nhe \: \: \: \: \: \: \: \: \: \:$
Gửi bởi ngoctruong236 trong 27-07-2013 - 17:32
$\dpi{150} \small \:Dat \:b _{n}=\frac{1}{u_{n}}\rightarrow \:b_{1}=2 \:va \: b_{n+1}=b_{n}^2-b_{n}+1\rightarrow \frac{1}{b_{n+1}-1}=\frac{1}{b_{n}(b_{n}-1)}=\frac{1}{b_{n}}-\frac{1}{b_{n}-1}\rightarrow \frac{1}{b_{n}}=\frac{1}{b_{n-1}}-\frac{1}{b_{n+1}-1}\ \:Ta \:có \::\sum_{i=1}^{n} u_{i}=u1+u2+...+un=\frac{1}{b_{1}}+\frac{1}{b_{2}}+.....+\frac{1}{b_{n}}= \frac{1}{b_{1}-1}-\frac{1}{b_{2}-1}+\frac{1}{b_{2}-1}-\frac{1}{b_{3}-1}+.....+\frac{1}{b_{n}-1}-\frac{1}{b_{n+1}-1}=\frac{1}{b_{1}-1}-\frac{1}{b_{n+1-1}}=\frac{1}{2-1}-\frac{1}{b_{n+1}-1}< 1\rightarrow dpcm\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$
Gửi bởi ngoctruong236 trong 27-07-2013 - 14:23
$\dpi{150} \small Áp\: dung\:bdt \:Cauchy \:cho \:3 \:số \:ko \:âm \:a,b,c. \:Ta có: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:ab^2=a.b.b\leq \frac{a^3+2b^3}{3} \:Tg \:tu, \:ta \:cung co \: bc^2\leq \frac{b^3+2c^3}{3},\:ca^2\leq \frac{c^2+2a^2}{3} \rightarrow ab^2+bc^2+ca^2\leq a^3+b^3+c^3.Áp dụng BDt Holder cho 3 so ta có:9(a^3+b^3+c^3)=(1^3+1^3+1^3)(a^3+b^3+c^3)(1^3+1^3+1^3)\geq (1.1.a+1.1.b+1.1.c)^3\rightarrow 9(a^3+b^3+c^3)\geq (a+b+c)^3,lai có abc\leq \frac{(a+b+c)^3}{27\rightarrow }\:ab^2+bc^2+ca^2+abc\leq a^3+b^3+c^3+abc\leq \frac{(a+b+c)^3}{9}+\frac{(a+b+c)^3}{27}=4\rightarrow dpcm \: \: \: \: \: \: \:$
Gửi bởi ngoctruong236 trong 27-07-2013 - 13:41
$\dpi{150} \small \:AB=2a\:\rightarrow AH=a. \: Ta\:có \:\OH^2=R^2-a^2\rightarrow \:OH=\sqrt{R^2-a^2}\rightarrow HE=OE-OH=R-\sqrt{R^2-a^2} \:Ta \: có\:AE^2=AH^2+HE^2=a^2+(R-\sqrt{R^2-a^2})^2=2R^2-2R\sqrt{R^2- a^2}+a^2=R^2-2R\sqrt{R^2-a^2}+R^2-a^2+2a^2=(R-\sqrt{R^2-a^2})^2+2a^2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$
Gửi bởi ngoctruong236 trong 25-07-2013 - 22:51
Gửi bởi ngoctruong236 trong 25-07-2013 - 21:52
minh nham thong cam nhegần nhất bạn ạ, vì mình xét hai nửa mặt phẳng cơ mà
Gửi bởi ngoctruong236 trong 25-07-2013 - 21:28
$\dpi{150} \small Ta co:x^6+y^4\geq 2\sqrt{x^6y^4}=2x^3y^2\rightarrow \frac{2x}{x^6+y^4}\leq \frac{1}{x^2y^2}\rightarrow \sum \frac{1}{x^2y^2}\geq \sum \frac{2x^6}{x^6+y^4}.Lai co \sum \frac{1}{x^2y^2}\leq \sum \frac{1}{x^4}(BDT Co si)\rightarrow dpcm$
Gửi bởi ngoctruong236 trong 25-07-2013 - 21:17
Gửi bởi ngoctruong236 trong 25-07-2013 - 20:32
Gửi bởi ngoctruong236 trong 25-07-2013 - 20:29
Gửi bởi ngoctruong236 trong 25-07-2013 - 19:24
Gửi bởi ngoctruong236 trong 25-07-2013 - 18:03
$\dpi{150} \small Khong mat tinh tong quat gia su a\leq b\leq c.Ta cóp^2=(a^2+b^2+c^2)^2\rightarrow \left [(a^4+b^4+c^4)+2(a^2b^2+b^2c^2+a^2c^2) \right ]\vdots p.Ma (a^4+b^4+c^4)\vdots p\rightarrow 2(a^2b^2+b^2c^2+a^2c^2)\vdots p.Mat\neq a,b,c\geq 1\rightarrow p\geq 3\rightarrow (p,2)=1\rightarrow (a^2b^2+b^2c^2+a^2c^2)\vdots p .Do đó \left [ a^2b^2+c^2(a^2+b^2+c^2)-c^4 \right ]\vdots p\rightarrow a^2b^2-c^4\vdots p\Leftrightarrow (ab-c^2)(ab+c^2)\vdots p.Lai có ab< 2ab\leq a^2+b^2\rightarrow 1< ab+c^2< a^2+b^2+c^2=p\rightarrow (ab+c^2,p)=1\rightarrow ab-c^2\vdots p.Mat\neq 1\leq a\leq b\leq c\rightarrow 0\leq c^2-ab< c^2< a^2+b^2+c^2=p\rightarrow c^2-ab=0\rightarrow c^2=ab\rightarrow p=3a^2.Ma p la số nguyên tố\rightarrow a=1\rightarrow p=3$
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