$\:Tim \:cac \:so \:tu \:nhien \:n/d(n)^{3}=4n \: \: \: \:$

$\Leftrightarrow 2 cos\frac{A+B}{2}cos\frac{A-B}{2}+1-2sin^2\frac{C}{2}\leq \frac{3}{2}\Leftrightarrow 4sin^2\frac{C}{2}-4sin\frac{C}{2}cos\frac{A-B}{2}+1\geq 0\Leftrightarrow (2sin\frac{C}{2}-cos\frac{A-B}{2})^2+sin^2\frac{A-B}{2}\geq 0\rightarrow dpcm.\:Dau \:bang \:xay \: ra\: \Leftrightarrow \Delta ABC\: deu$

$\;Xet \;dk \;thu \; 2:\;Voi \;n=1,2,3 \;thi \;m< \frac{1}{4} (thay\: vao\: BDT\:la \:dc )\;.Voi \;n=4 \; thi\; m=\frac{1}{4}.\; Nhu\; vay\; ta\;se \;chon \;m=\frac{1}{4} \;va \;diem \;xuat \; \;phat \;qui \; nap\;la \;n=4 \;Voi \;n=4 \;thay \;vao \;BDt \;ta \; dc\; 1066<1071(thoa man).\;Gia \;su \;BDT \;dg \;voi \;n=k\rightarrow S_{k}=\frac{1}{k+1}+\frac{1}{k+2}+.....+\frac{1}{2k} <\frac{7}{10}-\frac{1}{4k}\;,ta \; phai\;Cm \;BDT \;dg \;voi \; n=k+1\; \;hay S_{k+1} =\frac{1}{k+2}+\frac{1}{k+3}+....+\frac{1}{2k+2}< \frac{7}{10}-\frac{1}{4(k+1)}.Theo\;gt \;qui \;nap \;ta \;co \;S_{k+1}=S_{k}-\frac{1}{k+1} +\frac{1}{2k+1}+\frac{1}{2k+2}=Sk+\frac{1}{2(k+1)(2k+1)}< \frac{7}{10}-\frac{1}{4k}+\frac{1}{2(k+1)(2k+1)}.\;Nhu \; vay\;chi \;can \;CM \;\frac{-1}{4k}+\frac{1}{2(k+1)(2k+1)}< \frac{-1}{4(k+1)}\Leftrightarrow \frac{2}{(k+1)(2k+1)}< \frac{1}{k(k+1)}\Leftrightarrow 2k<2k+1\Leftrightarrow 0<1\rightarrow Bai \;toan \;dc \;CM \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;$

$\;Dat \;S_{k}=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n} \;. \; Neu\;giai \;bai \; toan\; bang\;p^2 \; qui\; nap\; thong\;thuong \;thi \;kho \;ma \;giai \;dc \;. \; Ta\;se \; tim\;1 \;so \;thuc \; m/BDT\;sau \; dung:\;\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\frac{1}{2n}<\frac{7}{10}-\frac{m}{n} \; .So\; m\;phai \;thoa \; man\; 2\;dk: \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (+)\;Buoc \;chuyen \;qui \;nap \;tu \;k \;sang \; k+1\;phai \; lam\; dc\; \; \; \; \; \; \; \; \(+); \;BDT \; tren\; phai\;dung \; voi\; gia\; tri\; dau\;cua \;n(co\:the \:\neq gia \:tri \:dau\:cua \:BDT \\:de \: bai ) \;.Xet \;dk \;1,ta \; co:S_{k+1}=Sk+\frac{1}{2(k+1)(2k+1)}< \frac{7}{10}-\frac{m}{k}+\frac{1}{2(k+1)(2k+1)}\Leftrightarrow \frac{1}{2(k+1)(2k+1)}+\frac{-m}{k}< \frac{-m}{k+1}\Leftrightarrow \frac{1}{2(k+1)(2k+1)}< \frac{m}{k(k+1)} \Leftrightarrow \;2m(2k+1)>k \Leftrightarrow (4m-1)k+2m>0\\;BDT \;cuoi \;nay \; dung\; voi\;moi \;k\Leftrightarrow m\geq \frac{1}{4} \; \; \; \; \; \; \; \; \; \; \; \;$

Đay chinh la BDT Holder cho 3 so ma ban

$\; Ta\;co \;:a^2+b^2+c^2=1\rightarrow 0\leq a,b,c\leq 1 , \;vi \; k\;la \;so \;nguyen \;duong \;chan \;\rightarrow k=2n(n>0) \rightarrow a^k=a^{2n}= (a^2)^n\leq a^2,\; tuong\; tu\; b^k\leq b^2,c^k\leq c^2\rightarrow f(a,b,c)\geq \frac{a}{b^2+c^2}+\frac{b}{a^2+c^2}+\frac{c}{a^2+b^2}= \frac{a}{1-a^2}+\frac{b}{1-b^2}+\frac{c}{1-c^2 }=\frac{a^2}{a(1-a^2)}+\frac{b}{b(1-b^2)}+\frac{c}{c(1-c^2)}=A\; .Ap\;dung \;BDT \;Cauchy \;cho \;3 \; so\; duong\;,ta \; co\; \sqrt[3]{2a^2(1-a^2)(1-a^2)}\leq \frac{2a^2+1-a^2+1-a^2}{3}=\frac{2}{3}\rightarrow a^2(1-a^2)^2\leq \frac{4}{27}\rightarrow \frac{a^2}{a(1-a^2)}\geq\frac{3\sqrt{3}}{2}a^2,tg \; tu; \;cho \;b \;va \; c\rightarrow A\geq \frac{3\sqrt{3}}{2}(a^2+b^2+c^2)= \frac{3\sqrt{3}}{2}\;.Vay \;min \;f(a,b,c)= \frac{3\sqrt{3}}{2} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;$

Bài 7 :

Bất đẳng thức cần chứng minh tương đương với:

$\frac{c(a-b)}{b(b+c)} +\frac{a(b-c)}{c(c+a)}+\frac{b(c-a)}{a(a+b)}\geq 0\; \; \; \;. \;$

Không mất tính tổng quát giả sử:$b$ nằm giữa $a$ và $c$;

Khi đo: $(b-a)(b-c)\leq 0 $\

Mặt khác: $b(c-a)=-c(a-b)-a(b-c)$

Do đó: Bất đẳng thức cần chứng minh tương đương với:

$c(a-b)\left [ \frac{1}{b(b+c)}-\frac{1}{c(a+b)} \right ]+a(b-c)\left [ \frac{1}{c(c+a)}-\frac{1}{a(a+b)} \right ]\geq 0$$

\Leftrightarrow \frac{c\left [ (a-b^2(a+b)+b(a-b)(a-c) \right ]}{(a+b)ab(b+c)}+\frac{\left [ (b-c)(a-c)(a+c)+a(b-c)^2 \right ]}{c(c+a)(a+b)}\geq 0 $

Mặt khác:$ (a-b)(a-c)=(a-b)^2-(b-a)(b-c)\geq 0,(b-c)(a-c)=(b-c)^2-(b-a)(b-c)\geq 0\Leftrightarrow dpcm\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;$

dg dinh lam thi bangbang lam nhanh qua

$\dpi{150} \:Day \:la \: cach\:lam \:cau \: c)\: cua\:minh. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:Goi \:J \: la\: giao\: diem\:cua \: XT\:va \:YZ, \:theo \: dly\: \:Tales \: co\:: \:\frac{IT}{IC}= \frac{MT}{MB}=\frac{ZT}{AB}= \frac{ZT}{CD}= \frac{TJ}{CO}\(do\Delta JTZ\: va\: \Delta OCD\:la \:2\Delta \:vuong \:can \: ): \: \: \: \: \: \; \; \; \;Mat\neq TJ \;song \; song\;voi \;CO\rightarrow T,J,O \; thang\; hang\;\rightarrow XT,YZ,OI dong \; qui\; \; \; \; \; \; \;Goi \; H\;la \;giao \; diem\; EM\;voi \; AB\;(E\: la\:giao \:cua \:dt \:qua \: M\: vuong\:goc \: voi\:CD \:va \:duong \:tron ),ta \;co \;\frac{IJ}{IO}=\frac{IT}{IC}=\frac{MT}{MB}=\frac{MI}{MH}=\frac{IK}{IE}(de\:y \:rang \: K\:va \: M\:dx\:nhau \:qua \:CD \: \: ) \;.Vay \; JK\;song \;song \;voi\:OE \:\rightarrow \frac{JK}{OE} = \frac{IJ}{IO}= \frac{JT}{OC}\:,ma \;OE=OC\rightarrow DPCM$

$\dpi{150} \:Gia \:su \:(a,b)=d\rightarrow a=k1d,b=k2d\rightarrow 2^a-1=2^{k1d}-1=(2^d-1).A,2^b-1=2^{k2d} -1=(2^d-1).B\rightarrow dpcm\: \: \: \: \: \: \: \: \: \: \: \: \:$

bai cuoi phai co them dkien m,n nguyen duong moi lam dc

$\dpi{150} \:Neu \: p=2\:thi \: moi\:n=2k \:deu \:thoa \:man \:. \:Gia \:su \p> 2,theo \:dly \:nho \:Fermat, \:ta \: co\: 2^{m(p-1)}\equiv 1(modp)\:.Lay \:n=m(p-1) \:voi \: m\equiv-1(mod p) \:\rightarrow n=m(p-1)\equiv 1(modp) \:va2^n-n\equiv 2^n-1\equiv 0(modp). \: \:Do \:co \:vo \:so \:so \: nguyen\: duong\:m \: sao \:cho \:m\equiv-1(mod p)\rightarrow ton \:tai \: vo\: so\: so\: nguyen\:duong \: n\: TMDK\:da \:cho\rightarrow dpcm \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\dpi{150} \:Gia \: su\: so\:nguyen \:to \ \:TMDK \:da\:cho\rightarrow 5^{2p^2} \equiv 1(mod p)\:.Mat\neq p^2-1\vdots p-1 ,\:theo \: dly\:nho \: Fermat\: ta\:co \: 5^{2(p^2-1)}\equiv 1(modp)\rightarrow 5^2\equiv 1(mod p)\:\rightarrow p=2hoac3 \: .Thu\:lai \:ta \:tim \:dc \=3 \: \: \: \: \: \: \:$

Phương trình đã cho tương đương với :

                                                          $x^{2}+y^{2}+(z+xy)^{2}-(xy)^{2}-1=0$

Hay $(x^{2}-1)(y^{2}-1)=(z+xy)^{2}$

Ta cần có $x^{2}-1=r^{2}.q$ và $y^{2}-1=s^{2}.q$

Hai phương trình này tương đương với 2 phương trình pell ; mà phương trình pell có vô số nghiệm nên phương trình này cũng có vô số nghiệm .

ban giai giong minh the

Chứng minh phương trình $x^{2}+y^{2}+z^{2}+2xyz=1$ có vô số nghiệm nguyên.

$\dpi{150} \small \:Ban \:tu \: ve\: hinh\:nha \:. \: \: \: \: \: \: \: \: \: \: \: \:Bai \:lam \:cua \: minh:\:Goi \:tam \:duong \:tron \:ngoai \: tiep\:\Delta ABC\rightarrow OA=OB=OC=R\:.Ta \:de \:dang \:cm\:dc\::OA\ vuong\:goc \: voi\: KI,\:OB \: vuong\:goc \: voi\: KH\:, \:OC \:vuong \:goc \: voi\: HI\:(cai\:nay \:ban \:tu \:CM \:nhe \:(de lam )\: ) \: \rightarrow \: S\:tu \:giac \:OKAI,OKBH,OICH \:lan \: luot\:=R.KI,R.KH,R.IH \: \rightarrow S\Delta ABC=R(KI+KH+IH)=R.p\Delta HIK\: \:Goi \:tam \:duong \: tron\: noi \:tiep \: \Delta HIK\:la \: r\rightarrow \SDelta HIK=r.p.\Delta HIK\rightarrow \frac{S\Delta HIK}{S\Delta ABC}=\frac{r.p\Delta HIK}{R.p\Delta HIK}=\frac{r}{R}$

Ta có: BĐT tương đương

$\sum \frac{3a^{3}}{3(a^{2}+b^{2})+(a-b)^{2}}\geq \frac{a+b+c}{2}\Leftrightarrow \sum (a-\frac{3b^{2}a+a(a-b)^{2}}{3(a^{2}+b^{2})+(a-b)^{2}})\geq \frac{a+b+c}{2}\Leftrightarrow \sum \frac{3b^{2}a+a(a-b)^{2}}{3(a^{2}+b^{2})+(a-b)^{2}}\leq \frac{a+b+c}{2}\Leftrightarrow \sum \frac{b}{2}(1-\frac{6ab}{3(a^{2}+b^{2})+(a-b)^{2}})-\sum \frac{a(a-b)^{2}}{3(a^{2}+b^{2})+(a-b)^{2}}\geq 0\Leftrightarrow \sum (a-b)^{2}(\frac{2b-a}{3(a^{2}+b^{2})+(a-b)^{2}})\geq 0$

TH1: Giả sử $a\geq b\geq c$

Ta dễ dàng chứng minh được $(a-c)S_{b}+(a-b)S_{c}\geq 0,(a-c)S_{b}+(b-c)S_{a}\geq 0$,do $S_{a}+S_{c}\geq 0$,mà a-c $\geq a-b$ nên  $(a-c)S_{b}+(a-b)S_{c}\geq 0,còn (a-c)S_{b}+(b-c)S_{a}\geq 0$ $\Leftrightarrow (2ab+2c^{2}+4ac-5bc)(ab-c^{2})\geq 0$,đúng theo giả thiết.Đây là tiêu chuẩn 4 nên ta có đ.p.c.m

TH2:TH này khó hơn chút,giả sử $a\geq c\geq b$

Ta có ngay $S_{a},S_{b}\geq 0$

Chỉ cần chứng minh $S_{c}+S_{a}\geq 0\Leftrightarrow 2abc+2b^{3}+4bc^{2}+2a^{2}c+2b^{2}c\geq ab^{2}+2ac^{2}+2a^{2}b+3abc$

Lại có $2abc+2a^{2}c\geq 2ac^{2}+2a^{2}b$ và $a\geq 2b\Rightarrow 2a^{2}c\geq b^{2}+3abc\Rightarrow$ nên suy ra $S_{a}+S_{c}\geq 0$,theo tiêu chuẩn 1 ta có đ.p.c.m

Từ đây chứng minh được bài toán,đẳng thức xảy ra khi và chỉ khi a=b=c

dg y toi đó

$\dpi{150} \small \:Ta \:co \::x1+x2=-7,x1.x2=1 \: \rightarrow x1\: va\:x2 \: khong\:chia \:het \:cho \:7 \:.Do \: do\: ap\: dung\:dinh \: ly\:Fecma \:, \:ta \:co \:x1^6\equiv 1(mod7) \:\rightarrow (x1^6)^{335}\equiv 1(mod7)\:\rightarrow x1^{2010}\equiv 1(mod7) \: Tg\:tu \:x2^{2010}\equiv 1(mod7) \:\rightarrow x1^{2010}+x2^{2010}\equiv 2(mod7) \:\rightarrow \:x1^{2010}+x2^{2010}=7k+2\rightarrow \:(x1^{2010}+x2^{2010})(x1^3+x2^3)=x1^{2013}+x2^{2013}+(x1x2)^3(x1^{2007}+x2^{2007})=(7k+2)(x1^3+x2^3)\: \:Ta \:co \:(x1x2)^3(x1^{2007}+x2^{2007}=-1(x1+x2)(x1^{2006}-x1^{2005}+.....+x2+1)=7(x1^{2006}-x1^{2005}+.....+x2+1)\vdots 7 \:,(7k+2)(x1^3+x2^3)\vdots 7 \:\rightarrow \:x1^{2013} +x2^{2013}=S2013\vdots 7\: \: \: \: \: \: \: \: \: \: \:$

Cho các số thực dương $a,b,c$. Chứng minh:

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

$\dpi{150} \small \: Theo\:dly \:Vi-et \: ta\:co \:: \:x1+x2=-7,x1.x2=-1 \:.Ta \:CM \:cau \:a \:theo \\:qui \:nap \: .\: Ta\: co\: :\:voi \:n=2 \: thi\: S2=(x1+x2)^2-2x1x2=49+2=51\:la \:so \: nguyen\: ,\: voi\:S3,S4 \: cung\:nhu \:vay \:. \:Gia \:su \: menh\: de\:dung \: voi\: n=k\:\rightarrow Sk \:la \:so \: nguyen\:.Menh \: de\: dung\: voi\:n=k \:\rightarrow \:menh \:de \: cung\:dung \:voi \:n=k-1 \:. \:Ta \: co\: Sk-1.Sk+1=(x1^{k-1}+x2^{k-1})(x1^{k+1}+x2^{k+1})=x1^k+x2^k+x1^{k-1}.x2^{k-1}(x1^2+x2^2)=VP\.:Theo \:qui \: nap\: \rightarrow \: VP\:la \:so \:nguyen \:\rightarrow Sk+1 \:la \:so \: nguyen\rightarrow dpcm$

$\dpi{150} \small \: Bai\: 1\: :Ta\:co: \:A= n^2+3n-38=(n-2)(n+5)-28\:. \:Do \:n+5-(n-2)=7\rightarrow \:hai \:so \:nay \:cung \:chia \:het \: cho\: 7\:hoac \:ca \:hai \:deu \:khong \:chia \:het \:cho \: 7\:\: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:Th1:n-2,n+5 \:cung \:chia \: het\:cho7\rightarrow (n-2)(n+5) \vdots 49,28\: khong\:chia \:het\:cho \:49\rightarrow A \:khong\vdots \:cho \:49 \:.Th2 \: CMTT\rightarrow dpcm\:$

Ta \; co\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;(m-n)^3+(m+n)^3=2m^3+6mn^2 \; \; \; \; \; \; \; \; \; \; \;Ta \;tim \;nghiem \;\; \;co \;dang \; (x,y,z,t)=(a-b,a+b,\frac{c}{2}-\frac{d}{2},\frac{c}{2}+\frac{d}{2})\; voi\;a,b,c,d \;la \;cac \;so \;nguyen \;.De \;thay \;(x,y,z,t)=(10,10,-1,0) \; la\;1 \;nghiem \;.Ta \; se\; tim\; nghiem\;voia=10 \;va \;c=1 co \; dinh\;. \;De \; thay\;(x,y,z,t)=(10-b,10+b,\frac{-1}{2}-\frac{d}{2},\frac{-1}{2}+\frac{d}{2}) \; la\;nghiem \;cua \;phuong \;trinh \;da \;cho\Leftrightarrow (2000+60b^2)-\frac{1+3d^2}{4} =1999\;hay \; \;d^2-80b^2=1 \;.De \;thay \; phuong\; trinh\:Pell \: d^2-80b^2=1\: co\:nghiem \:nguyen \:duong \:nho \:nhat \:la \:(d1,b1)=(9,1) \: .Do\:Pt \:Pell \:co \:vo \:so \:nghiem \: nen\:Pt \: da\:cho \:cung \:\:co \: vo\: so\:nghiem\\rightarrow dpcm \: \:

$\dpi{150} \small \: Ap\:dung \: BDT\: C-S,\:ta \: co\:: \:\frac{MB^2}{AB^2}+\frac{MC^2}{AC^2} \geq \frac{(MB+MC)^2}{AB^2+AC^2}\geq \frac{BC^2}{AB^2+AC^2}=1\rightarrow dpcm\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\dpi{150} \small \:PT \: \Leftrightarrow \:x^3-2x^2+2x^2-4x+x-2=\sqrt{x+2}-2\Leftrightarrow (x+1)^2(x-2)=\frac{x-2}{\sqrt{x+2}+2} \Leftrightarrow (x-2)\left [ (x+1)^2-\frac{1}{\sqrt{x+2}+2} \right ]=0\\.: Voi\: x<2 va x>2\:thi \:(x+1)^2-\frac{1}{\sqrt{x+2}+2} \:deu \: lon \: va\:nho \: hon\:0 \:\rightarrow x=2 \:Vay \:phuong \:trinh \: co\:1 \:no \:x=2 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\dpi{120} \small Ton \: tai\: day\left \{ x1,x2,.... xn\right \}=\left \{ 1,2,.....n \right \}\: thoa\: man\: x1+x2+.....+xk\vdots k \forall k=1,2.....n$