ta có \(x+y+z=2019xyz=>2019x^2=\frac{x^2+xy+xz}{yz}\)

\(=>2019x^2+1=\frac{x^2+xy+xz+yz}{yz}=\frac{\left(x+y\right)\left(x+z\right)}{yz}=\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)\)

\(=>\sqrt{2019x^2+1}=\sqrt{\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)}\le\frac{1}{2}\left(\frac{x}{y}+\frac{x}{z}+2\right)=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)

(theo BDT cô -si)

\(=>\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le\frac{x^2+1+1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=x+\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)

tương tự \(\frac{y^2+1+\sqrt{2019y^2+1}}{z}\le y+\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\)

\(\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le z+\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)

=>.vt\(\le x+y+z+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)

chứng minh được \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)

=>\(3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3\left(xy+yz+zx\right)}{2019xyz}\le\frac{2019\left(x+y+z\right)^2}{x+y+z}=2019\left(x+y+z\right)\)

=>.vt\(\le2020\left(x+y+z\right)=2020.2019xyz=\)vt

=> dpcm

Ta có: \(2019xyz=x+y+z\)

=> \(2019xy=\frac{x}{z}+\frac{y}{z}+1>1\); \(2019yz=\frac{y}{x}+\frac{z}{x}+1>1\); \(2019xz=\frac{x}{y}+\frac{z}{y}+1>1\)

Ta  lại có: \(x+y+z=2019xyz\)

=> \(2019x\left(x+y+z\right)=2019^2x^2yz\)

=> \(2019x^2+1=\left(2019^2x^2yz-2019xy\right)-\left(2019xz-1\right)\)

=> \(2019x^2+1=\left(2019xy-1\right)\left(2019xz-1\right)\le\frac{\left(2019xy+2019xz-2\right)^2}{4}\)

=> \(\sqrt{2019x^2+1}\le\frac{2019xy+2019xz-2}{2}\)

Tương tự : \(\sqrt{2019y^2+1}\le\frac{2019xy+2019yz-2}{2}\)

\(\sqrt{2019z^2+1}\le\frac{2019xz+2019yz-2}{2}\)

=> \(\frac{x^2+1+\sqrt{2019x^2+1}}{x}+\frac{y^2+1+\sqrt{2019y^2+1}}{y}+\frac{z^2+1+\sqrt{2019z^2+1}}{z}\)

\(\le\)\(\frac{x^2+1+\frac{2019xy+2019xz-2}{2}}{x}+\frac{y^2+1+\frac{2019xy+2019yz-2}{2}}{y}+\frac{z^2+1+\frac{2019xz+2019yz-2}{2}}{z}\)

\(=\frac{2x^2+2019xy+2019xz}{2x}+\frac{2y^2+2019xy+2019yz}{2y}+\frac{2z^2+2019xz+2019yz}{2z}\)

\(=x+\frac{2019}{2}y+\frac{2019}{2}z+y+\frac{2019}{2}x+\frac{2019}{2}z+z+\frac{2019}{2}x+\frac{2019}{2}y\)

\(=2020\left(x+y+z\right)=2020.2019xyz\)

Vậy có điều cần cm

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x=y=z\\x+y+z=2019xyz\end{cases}}\Leftrightarrow x=y=z=\frac{1}{\sqrt{673}}\)

M A C D O B N P Q E

Dễ thấy \(\Delta MCB~\Delta MDC\left(g.g\right)\Rightarrow\frac{MC}{MD}=\frac{BC}{CD}\)( 1 )

\(\Delta MAB~\Delta MDA\left(g.g\right)\Rightarrow\frac{MA}{MD}=\frac{AB}{AD}\)( 2 )

Lại có MA = MC . Từ ( 1 )  và ( 2 ) suy ra \(\frac{BC}{CD}=\frac{AB}{AD}\Rightarrow AD.BC=AB.CD\)

Áp dụng định lí Ploleme với tứ giác ABCD, ta có :

\(AB.CD+AD.BC=AC.BD\)

\(\Rightarrow BC.AD=AC.BD-AB.CD=\frac{1}{2}AC.BD\)

\(\Rightarrow\frac{AC}{AD}=\frac{2BC}{BD}\)( 3 )

\(\Delta NBE~\Delta NDB\left(g.g\right)\Rightarrow\frac{NB}{ND}=\frac{BE}{DB}\); \(\Delta NCE~\Delta NDC\left(g.g\right)\Rightarrow\frac{NC}{ND}=\frac{CE}{CD}\)

lại có :  NB = NC \(\Rightarrow\frac{BE}{BD}=\frac{CE}{CD}\Rightarrow BE.CD=CE.BD\)

Áp dụng định lí Ptoleme với tứ giác BECD, ta có : 

\(BE.CD+CE.BD=BC.DE\Rightarrow BE.CD=CE.BD=\frac{1}{2}BC.DE\)

\(\Delta PBC~\Delta PDB\left(g.g\right)\Rightarrow\frac{PC}{PB}=\frac{PB}{PD}\Rightarrow PC.PD=PB^2\)

Mà \(\frac{PC}{PB}=\frac{PB}{PD}=\frac{BC}{BD}\)

Mặt khác : \(\frac{PC}{PD}=\frac{PC.PD}{PD^2}=\left(\frac{PB}{PD}\right)^2=\left(\frac{BC}{BD}\right)^2\)( 4 )

suy ra : \(\frac{PC}{PD}=\left(\frac{BC}{BD}\right)^2=\left(\frac{2CE}{DE}\right)^2\)

giả sử AE cắt CD tại Q

\(\Rightarrow\Delta QEC~\Delta QDA\left(g.g\right)\Rightarrow\frac{QC}{QD}=\left(\frac{2CE}{DE}\right)^2\)

\(\Rightarrow\frac{QC}{QD}=\frac{PC}{PD}\Rightarrow P\equiv Q\)

Vậy 3 điểm A,E,P thẳng hàng

v mình quên nối AE cắt CD. hay là nối 3 điểm A,E,P mà thôi, không sao.

vì b,c là nghiệm của phương trình nên \(\hept{\begin{cases}b^2-ab-\frac{1}{2a^2}=0\\c^2-ab-\frac{1}{2a^2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}b^4=\left(ab+\frac{1}{2a^2}\right)^2\\c^4=\left(ac+\frac{1}{2a^2}\right)^2\end{cases}}\)

\(b^4+c^4=\left(ab+\frac{1}{2a^2}\right)^2+\left(ac+\frac{1}{2a^2}\right)^2\ge\frac{1}{2}\left(ab+ac+\frac{1}{a^2}\right)^2\)

\(=\frac{1}{2}\left[a\left(b+c\right)+\frac{1}{a^2}\right]^2\)

mà theo viet : (tính delta đầu tiên nhá ): b+c=a.

\(\Rightarrow b^4+c^4\ge\frac{1}{2}\left(a^2+\frac{1}{a^2}\right)^2\ge2\)(AM-GM)

Dấu = xảy ra khi a=1 hoặc a=-1

^^ Lời giải hay 

hihi em mới học lớp 3 thôi

đề bài sai rồi bạn nhé check lại đi 

Sửa đề: \(\frac{a}{b}+\frac{a}{c}+\frac{c}{b}+\frac{c}{a}+\frac{b}{c}+\frac{b}{a}\ge\sqrt{2}\left(\Sigma\sqrt{\frac{1-a}{a}}\right)\)

or \(\Sigma\frac{b+c}{a}\ge\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\)

Theo AM-GM:\(\frac{b+c}{a}\ge2\sqrt{\frac{2\left(b+c\right)}{a}}-2\)

Tương tự và cộng lại: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-6\)

Mà: \(\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}\ge3\sqrt[6]{\frac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}}\ge6\)

Từ đó: \(VT\ge2\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}-\Sigma\sqrt{\frac{2\left(b+c\right)}{a}}=VP\)

Done!

\(PT\Leftrightarrow5x^2+x\left(5y-7\right)+5y^2-14y=0\)

\(\Delta=\left(5y-7\right)^2-4.5.\left(5y^2-14y\right)\)

   \(=196-3\left(5y-7\right)^2\)

Để phương trình có nghiệm thì \(\Delta\ge0\Rightarrow\left(5y-7\right)^2\le65\)

Mặt khác \(5y-7\equiv3\left(mod5\right)\)

\(\Rightarrow\left(5y-7\right)^2\equiv4\left(mod5\right)\)

do đó \(\left(5y-7\right)^2\in\left\{4,9,14,19,24,29,34,39,44,49,54,59,64\right\}\)

mà (5y-7)² là số chính phưng nên \(\left(5y-7\right)^2\in\left\{4,9,64\right\}\)

Từ đó tính ra

\(5\left(x^2+xy+y^2\right)=7\left(x+2y\right)\)

\(\Leftrightarrow5x^2+5xy+5y^2-7x-14y=0\)

\(\Leftrightarrow5x^2+x\left(5y-7\right)+5y^2-14y=0\)

\(\Rightarrow\Delta_x=\left(5y-7\right)^2-4\cdot5\cdot\left(5y^2-14y\right)\)

\(=-75y^2+210y+49\)

\(=196-3\left(25y^2-2\cdot5y\cdot7+79\right)\ge0\)

\(=196-3\left(5y-7\right)^2\ge0\)

Để phương trình có nghiệm nguyên thì \(\Delta_x\ge0\Leftrightarrow\left(5y-7\right)^2\le65\)

Nhận thấy \(5y-7\equiv3\left(mod5\right)\Rightarrow\left(5y-7\right)^2\equiv4\left(mod5\right)\)

Do đó \(\left(5y-7\right)^2\in\left\{4;9;14;19;24;29;34;39;44;49;54;59\right\}\)

Mà \(\left(5y-7\right)^2\)chinh phương nên \(\left(5y-7\right)^2\in\left\{4;9;49\right\}\)

Đến đây ta xét trường hợp là ra.

Ta có \(\left(a-1\right)^2\left(a^2+a+1\right)\ge0\)\(\Leftrightarrow\left(a^2-2a+1\right)\left(a^2+a+1\right)\ge0\)

\(\Leftrightarrow a^4-a^3-a+1\ge0\)

\(\Leftrightarrow a^4-a^3+1\ge a\)

\(\Leftrightarrow a^4-a^3+ab+2\ge a+ab+1\)

\(\Rightarrow\frac{1}{\sqrt{a^4-a^3+ab+2}}\le\frac{1}{\sqrt{ab+a+1}}\)

Tương tự \(\frac{1}{\sqrt{b^4-b^3+bc+2}}\le\frac{1}{\sqrt{bc+b+1}}\)

             \(\frac{1}{\sqrt{c^4-c^3+ca+2}}\le\frac{1}{\sqrt{ca+c+1}}\)

Cộng từng vế các bđt trên ta được

\(VT\le\frac{1}{\sqrt{ab+a+1}}+\frac{1}{\sqrt{bc+b+1}}+\frac{1}{\sqrt{ca+c+1}}\)

Áp dụng bđt Bunhiacopski ta có

\(VT\le\sqrt{3\left(\frac{1}{ab+a+1}+\frac{1}{bc+b+1}+\frac{1}{ca+c+1}\right)}\)\(=\sqrt{3\left(\frac{1}{ab+a+1}+\frac{a}{abc+ab+a}+\frac{ab}{a^2bc+abc+ab}\right)}=\sqrt{3}\)

Dấu "=" xảy ra khi a=b=c=1

Đoán xem

Từ \(\frac{a}{1+a}+\frac{2b}{1+b}+\frac{3c}{1+c}+\frac{5d}{1+d}\le1\)

\(\Rightarrow1-\frac{a}{1+a}+2-\frac{2b}{1+b}+3-\frac{3c}{1+c}+5-\frac{5d}{1+d}\ge10\)

\(\Rightarrow\frac{1}{1+a}+\frac{2}{1+b}+\frac{3}{1+c}+\frac{5}{1+d}\ge10\)

Áp dụng BĐT AM-GM ta có: 

\(\frac{1}{a+1}\ge\)\(\frac{2b}{1+b}+\frac{3c}{1+c}+\frac{5d}{1+d}\ge10\sqrt[10]{\frac{b^2c^3d^5}{\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^5}}\)

Và \(\frac{1}{1+b}\ge\)\(\frac{a}{1+a}+\frac{b}{b+1}+\frac{3c}{c+1}+\frac{5d}{d+1}\)

\(\ge10\sqrt[10]{\frac{abc^3d^5}{\left(1+a\right)\left(1+b\right)\left(1+c\right)^3\left(1+d\right)^5}}\)

Và \(\frac{1}{1+c}\ge\frac{a}{1+a}+\frac{2b}{b+1}+\frac{2c}{c+1}+\frac{5d}{d+1}\)

\(\ge10\sqrt[10]{\frac{ab^2c^2d^5}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^2\left(1+d\right)^5}}\)

Và \(\frac{1}{1+d}\ge\frac{a}{a+1}+\frac{2b}{b+1}+\frac{3c}{c+1}+\frac{4d}{d+1}\)

\(\ge10\sqrt[10]{\frac{ab^2c^3d^4}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^4}}\)

Nhân theo vế 4 BĐT có: \(\frac{1}{\left(1+a\right)\left(1+b\right)^2\left(1+c\right)^3\left(1+d\right)^5}\)

\(\ge10^{1+2+3+5}\sqrt[10]{\frac{a^{2+3+5}b^{2+2+6+10}c^{3+6+6+15}d^{5+10+15+20}}{\left(1+a\right)^{10}\left(1+b\right)^{20}\left(1+c\right)^{30}\left(1+d\right)^{50}}}\)

Tương đương với \(ab^2c^3d^5\le\frac{1}{10^{11}}\) (ĐPCM)

kho ko

a, Ta co 2 bo de quen thuoc sau : FC la phan giac ^EFD, FB la phan giac PFD

ma QR//EP nen

\(\widehat{PFB}=\widehat{FQD}=\widehat{QFD}\Rightarrow\Delta DFQ\) can tai D => DF=DQ (1)

mat khac theo tinh chat tia phan giac ngoai ^PFD co \(\frac{FD}{FP}=\frac{CD}{CP}\) 

ma \(\frac{CD}{CP}=\frac{DT}{PF}\) (DT//PF)

suy ra \(\frac{DF}{PF}=\frac{DT}{PF}\Rightarrow DT=DF\) (2)

Tu(1)va (2) suy ra DT=DQ hay D la trung diem QT

b, Goi S la trung diem BC ta chung minh PQSR noi tiep 

Co \(\Delta PSE~\Delta ESD\left(G-G\right)\Rightarrow\frac{PS}{ES}=\frac{ES}{SD}\Leftrightarrow ES^2=PS.DS\)

lai co ES=SB=SC do S la trung diem canh huyen BC cua tam giac vuong BEC

suy ra \(BS^2=PS.SD=DS\left(PD+DS\right)=SD^2+PD.DS\)

=> \(PD.DS=BS^2-SD^2=\left(BS-DS\right)\left(BS+DS\right)=BD.DC\) (3)

Mat khac ^DQB=^PFB(cmt)

^PFB=^RCD( BFEC nt)

suy ra ^DQB=^RCD=> BQCR noi tiep

=> \(BD.DC=DQ.DR\) (4)

Tu (3),(4) suy ra DP.DS=DQ.DR => PQDR noi tiep 

=> (PQR) di qua S la trung diem BC co dinh

c,lay H' doi xung voi H qua BC, ta co H' thuoc (O) .

ta lai co bo de sau : \(BD.DC=DH.DA\) (quen thuoc)

suy ra \(DP.DS=DH.DA\left(=DB.DC\right)\)

<=> \(\frac{DH}{DP}=\frac{DS}{DA}\)

ma ^HDP=^SDA=90

suy ra \(\Delta DHP~\Delta DSA\left(c-g-c\right)\Rightarrow\widehat{DHP}=\widehat{DSA}\)

va \(\widehat{DSA}=\widehat{AHK}\left(phu\widehat{DAS}\right)\)

=>\(\widehat{DHP}=\widehat{AHK}\) => P,H,K thang hang

lai co \(\widehat{AFH}=\widehat{AKH}=\widehat{AEH}=90\)

=> A,F,H,K,E cung thuoc 1 duong tron =. FHKE noi tiep

=>\(PF.PE=PH.PK\) (5)

ma BFEC noi tiep => \(PF.PE=PB.PC\) (6)

(5)+(6)Suy ra \(PH.PK=PB.PC\) => BHKC noi tiep

Vi H' ,I doi xung voi H,K qua BC ma BHKC noi tiep => BH'IC noi tiep

do vay \(I\in\left(BH'C\right)=\left(ABH'C\right)=\left(O\right)\)

e,Goi tam (CJL) la U, (U) cat (O) tai V, BC giao OG tai X

=> \(\widehat{VBG}=\widehat{VJG}\left(=\widehat{VCB}\right)\) =>BJVG noi tiep

=> B,J,X,V,G cung thuoc 1 duong tron => ^BVG=^BXG=90

lai co ^XVG +^XBG=180 hay ^XVG+^BAC=180

va ^BVC+^BAC=180

suy ra ^XVG=^BVC

hay 90 +^XVB=^XVB+^XVC

=> ^XVC=90

=> V thuoc duong tron dk XC

mat khac V cung thuoc (O)

suy ra V co dinh ,C co dinh 

suy ra tam U di chuyen tren trung truc VC co dinh (dpcm)

em mới lớp 5 lên ko bít bài này