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} $

$\dpi{150} \small Ta co:x^6+y^4\geq 2\sqrt{x^6y^4}=2x^3y^2\rightarrow \frac{2x}{x^6+y^4}\leq \frac{1}{x^2y^2}\rightarrow \sum \frac{1}{x^2y^2}\geq \sum \frac{2x^6}{x^6+y^4}.Lai co \sum \frac{1}{x^2y^2}\leq \sum \frac{1}{x^4}(BDT Co si)\rightarrow dpcm$

$\;Ap \;dung \;BDT \;Cauchy \; cho\;3 \;so, \; ta\; co:\; 2(\frac{a}{a+b})^3+\frac{1}{8}=(\frac{a}{a+b})^3+\(\frac{a}{a+b})^3+\frac{1}{8}\geq 3\sqrt[3]{(\frac{a}{a+b})^3.(\frac{a}{a+b})^3.\frac{1}{8}}=\frac{3}{2}(\frac{a}{a+b})^2\rightarrow (\frac{a}{a+b})^3\geq \frac{3}{4}(\frac{a}{a+b})^2-\frac{1}{16}; \;.Tuong \; tu\;cho \;b \; va\;c\rightarrow \sum (\frac{a}{a+b})^3 \geq \frac{3}{4}\sum (\frac{a}{a+b})^2-\frac{3}{16}.Mat\;\neq \;ta \;co \;sum (\frac{a}{a+b})^2\geq \frac{3}{4}(de\,dang \,cm \,dc ) \rightarrow \;\sum (\frac{a}{a+b})^3\geq \frac{3}{8}\rightarrow dpcm \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;$

hinh nhu ban nham đe

phai la 2010 chứ

$\dpi{150} \small Dat \sqrt[3]{12-x}=a,\sqrt[3]{14+x}=b.Ta co:\left\{\begin{matrix} & a+b=2(1) & \\ & a^3+b^3=26(2) & \end{matrix}\right.(1)\rightarrow a=2-b,thay vao (2),ta co:(2-b)^3+b^3=26\Leftrightarrow 8-12b+6b^2=26\Leftrightarrow 6b^2-12b-18=0\Leftrightarrow 6(b-3)(b+1)=0\rightarrow b=3, a=-1 hoac b=-1,a=3\rightarrow x=13 hoac x=-15$

BAI 1 phai la n khac 0 moi dc

Đay chinh la BDT Holder cho 3 so ma ban

$\;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} \; \; \; \; \; \; \; \; \; \; \; \;$

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

roi danh gia trong doan 0 ,1/2 la ra

$Đặt P la biểu thức ở vế trái cua BDT cần cm Áp dụng BDT \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z},với mọi x,y,z>0 ta có: P=\frac{1}{x(1-x)}+\frac{1}{y(1-y)}+\frac{1}{z(1-z)}=(\frac{1}{x}+\frac{1}{1-x})+(\frac{1}{y}+\frac{1}{1-y})+(\frac{1}{z}+\frac{1}{1-z})=(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})+(\frac{1}{1-x}+\frac{1}{1-y}+\frac{1}{1-z})\geq \frac{9}{x+y+z}+\frac{9}{3-(x+y+z)} Chú y rang $0\leq x,y,z\leq \frac{1}{2}.Đăt S=x+y+z ta co P\geq \frac{9}{S}+\frac{9}{3-S}=\frac{27}{S(3-S)}= \frac{27}{\frac{9}{4}-(\frac{3}{2}-S)^2}.

ta có$(a^{3}+b^{2}+c)(\frac{1}{a}+1+c)\geq (a+b+c)^{2}\rightarrow \frac{a}{a^{3}+b^{2}+c}\leq \frac{1+a+ac}{(a+b+c)^{2}}$ tương tự và chú ý rằng $a^{2}+b^{2}+c^{2}\geq a+b+c$

cach nay hay that

$ap dụng bdt cauchy-schward ta coVT\geq \frac{a^2+b^2+c^2}^2{\sum a^3+\sum 2a^2b^2}.Cần Cm Bdt\geq 1\rightarrow phai cm \sum a^4\geq \sum a^3.vi a+b+c=3\rightarrow dpcm$

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

$\dpi{150} \small He pt\Leftrightarrow \left\{\begin{matrix} & 8x^{3}y^{3}+27=18y^{3}(1) & \\ & 4x^{2}y^2+6xy=y^{3}(2) & \end{matrix}\right.Chia (1) cho (2) ta dc \frac{8x^3y^3+27}{4x^2y^2+6xy}=18.Đat xy=a\rightarrow 8a^3+27=72a^2+108a\Leftrightarrow 8a^3-72a^2-108a+27=0\Leftrightarrow 8a^2(a+\frac{3}{2})-84a(a+\frac{3}{2})+18(a+\frac{3}{2}\Leftrightarrow (8a^2-84a+18)(a+\frac{3}{2})=0\Leftrightarrow \left\{\begin{matrix} & a+\frac{3}{2}=0(3) & \\ & 8a^2-84a+18=0(4) & \end{matrix}\right.(3\rightarrow a=\frac{-3}{2})\rightarrow xy=\frac{-3}{2}.Thay vao (1)\rightarrow y=0(loai).(4)\rightarrow 4a^2-42a+9=0.Tim ra la xong nhe$

$Coi phương trình 1 la phuong trinh bac hai an x , ta co: \Delta (1)=4-4y^4\geq 0\rightarrow -1\leq y\leq 1 Phương trinh (2)\Leftrightarrow 2(x-1)^2+1+y^3=0 Do y\geq 1\rightarrow y^3+1\geq 0,lai co 2(x-1)^2\geq \rightarrow 2(x-1)^2=0 va y^3+1=0\rightarrow x=1,y=-1$

$\dpi{200} \small Ta có x=y=z=0 ko phai la1 no cua he phuong trinh.Xét xyz\neq 0 Ta có:Đây là hệ hoán vị vòng quanh,dễ cm dc (x,y,z)la no thi x=y=z. Vậy x=y=z=\frac{-8}{7}$

$\dpi{150} \small Dat a=\overline{xy},b=\overline{ztu}.Tu gt suy ra:(a+b)^2=1000a+b\Leftrightarrow (a+b)(a+b-1)=999a(1).Lai co (a+b)^2\leq 9999\rightarrow a+b\leq 316.Ket hop voi (1)\rightarrow (a+b)\vdots 27 va (a+b-1)\vdots 37 hoac (a+b)\vdots 37 va (a+b-1)\vdots 27. TH1:a+b=27m,a+b-1=37n\rightarrow 27m=37n+1.GIai ta dc m=37k+11,n=27k+8(k\geq 0).THAy vao (1) ta dc a=mn=(37k+11)(27k+8)(k\geq 0).Vi 10\leq a\leq 99 nen k=0.khi do a=88,b=209 va\overrightarrow{xyztu}=88209.TH2: ta co 37m=27n+1 giai ta dc m=27k-8 va n=37k-11 THAy vao (1) ta dc a=mn=(27k-8)(37k-11)vi k\geq 1\rightarrow a\geq 19.26> 100(loai).Vay x=y=8,z=2,t=0,u=9$

bai toan nay con len dc bao ba co

$\inline \dpi{150} \small Ta co phuong trinh\Leftrightarrow x^3-2x^2-3x^2+6x+3x-6=\sqrt{2x-3}-1\Leftrightarrow (x-2)(x^2-3x+3)=\frac{2x-3-1}{\sqrt{2x-3}+1}\Leftrightarrow (x-2)\left [ (x^2-3x+3)-\frac{2}{\sqrt{2x-3}+1} \right ]=0\Leftrightarrow x=2.$

$\Leftrightarrow \overline{ab}^{3}=\overline{abcde}$$\Leftrightarrow \overline{ab}^{3}=\overline{abcde}$$\Leftrightarrow $Đặt\overline{ab}=x,\overline{cde}=y (10\leq x\leq 99),100\leq y\leq 999$$ta co x^{3}=1000x+y,y\geq 0\rightarrow x^{3}\geq 1000x\rightarrow x(x^{2})-1000)>0\rightarrow x^{2}> 1000\rightarrow x> 31$$.Tip theo ta co y< 1000\rightarrow x^{3}< 1000x+1000\rightarrow x^{3}-1000x< 1000\rightarrow x(x^{2}-1000)< 1000.Neu x\geq 33\rightarrow x(x^{2}-1000)\geq 1000\rightarrow x< 33\rightarrow x=32$$vay ta dc so can tim la 32768 va ab=32$

$\sqrt{\frac{b+c}{a}} +\sqrt{\frac{c+a}{b}}+\sqrt{\frac{a+b}{c}}\geq \sqrt{\frac{16(a+b+c)^{3}}{3(a+b)(b+c)(a+c)}}$

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

$\frac{\sqrt{a+b}}{c}+\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$

$\bigtriangleup ABC$ nt(O,r).các đường cao $AD,BE,CF$.$H$ la trực tâm.CM

a)$(AH+BH+CH)^2\leq a^2+b^2+c^2$ ($a,b,c$ la 3 canh tam giac)

b) $a^2+b^2+c^2\geq 8\sqrt{3}p{R}'$ ($p$ là nửa chu vi tam giac $DEF$,${R}')$ la bán kính $(DEF)$

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\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;$