ngoctruong236

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

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} $

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

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 \: \:

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

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

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

$\dpi{200} \small CMR:a\sqrt{\frac{b}{a+c}}+b\sqrt{\frac{c}{a+b}}+c\sqrt{\frac{a}{b+c}}\leq \frac{3\sqrt{3}}{4}\sqrt{\frac{(a+b)(b+c)(c+a)}{a+b+c}}$

hoan toan co the gia su a+b+c=3 hoac abc =1 trong bai nay ma anh

$\dpi{120} \small \;Thuc \;chat \; day \;chinh \; la\;bai \;toan \;: \;Cho \;a,b,c \;la \;cac \;so \;thuc \; duong\; thoa\;man \;a+b+c=3 \;CMR: \frac{a}{b^{3}+2}+\frac{b}{c^{3}+2}+\frac{c}{a^{3}+2}\geq 1\;.Ta CM \; BDT\;nay \; nhu\;sau \;.BDt \; can\;CM \;tuong \;duong \;voi \;(\frac{a}{2}-\frac{a}{b^3+2})+(\frac{b}{2}-\frac{b}{c^3+2})+\left ( \frac{c}{2}-\frac{c}{a^3+2} \right )\leq \frac{1}{2} \;hay \;\frac{ab^3}{b^3+2}+\frac{bc^3}{c^3+2}+\frac{ca^3}{a^3+2}\leq 1 \;Su \;dung \; BDT\;AM-GM \;,ta \;co \;b\leq \frac{b^3+1+1}{3}=\frac{b^3+2}{3}\rightarrow \frac{ab^3}{b^3+2}= \frac{ab^2.b}{b^3+2}\leq \frac{ab^2}{3}\; \rightarrow \sum \frac{ab^3}{b^3+2}\leq\ \frac{ab^2+bc^2+ca^2}{3}\;.Bai \;toan \; quy\; ve\; CM\; ab^2+bc^2+ca^2\leq 3.\;Mat \neq abc=1\rightarrow can CM \; ab^2+bc^2+ca^2+abc\leq 4\;.BDT \;nay \;chinh \; la\; BDT\;co \;ban \;(a+b+c)^3\geq \frac{27}{4}(ab^2+bc^2+ca^2+abc)\rightarrow dpcm \; \; \; \; \; \; \;$

cha hiu sao no ko ra dc latex

$\dpi{150} \small \:Nhan \:2 \: ve\:cua \:Bdt \:cho \:2 \:, \: ta\:can CM \:\frac{a+b}{ab}+\frac{b+c}{bc}+\frac{a+c}{ac}+\frac{2}{\sqrt[3]{abc}}\geq \frac{4(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}. \:Ap \:dung \:Bdt AM-GM, \:ta \:co \:: \:VT=\frac{(a+b)^2}{ab(a+b)}+\frac{(b+c)^2}{bc(b+c)}+\frac{(a+c)^2}{ac(a+c)}+\frac{(2\sqrt[3]{abc})^2}{2abc}\geq \frac{\left [ (a+b+b+c+c+a+2\sqrt[3]{abc}) \right ]^2}{ab(a+b)+bc(b+c)+ca(c+a)+2abc)}=\frac{4(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}= VP\rightarrow dpcm : \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

Bài này mình xin trình bày theo cách trâu bò húc $1$ tí

Quy đồng mẫu thức ta có bất đẳng thức tương đương với

       $\frac{\sum (b+c)(c+a}{(a+b)(b+c)(c+a)}+\frac{1}{2\sqrt[3]{abc}}\geqslant \frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}$

$\Leftrightarrow \frac{1}{2\sqrt[3]{abc}}\geqslant \frac{(a+b+c+\sqrt[3]{abc})^2-\sum (b+c)(c+a)}{(a+b)(b+c)(c+a)}$

Khai triển ta được $\Leftrightarrow \frac{1}{2\sqrt[3]{abc}}\geqslant \frac{\sqrt[3]{(abc)^2}+2\sqrt[3]{abc}(a+b+c)-(ab+bc+ca)}{(a+b)(b+c)(c+a)}$

       $\Leftrightarrow (a+b)(b+c)(c+a)\geqslant 2abc+4\sqrt[3]{(abc)^2}(a+b+c)-2\sqrt[3]{abc}(ab+bc+ca)$

       $\Leftrightarrow ab(a+b)+bc(b+c)+ca(c+a)+2\sqrt[3]{abc}(ab+bc+ca)\geqslant 4\sqrt[3]{(abc)^2}(a+b+c)$

D0 bất đẳng thức trên là thuần nhất nên ta có thể chuẩn hóa $abc=1$

Từ đó ta chỉ cần chứng minh bất đẳng thức sau

           $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}+2(ab+bc+ca)\geqslant 4(a+b+c)$ với $abc=1$

Áp dụng AM-GM ta có $2ab+\frac{a}{b}+\frac{b}{a}=(ab+\frac{a}{b})+(ab+\frac{b}{a})\geqslant 2a+2b=2(a+b)$

                                    $2bc+\frac{b}{c}+\frac{c}{b}\geqslant 2(b+c)$

                                    $2ca+\frac{c}{a}+\frac{a}{c}\geqslant 2(c+a)$

Cộng 3 bđt trên lại ta có đpcm

Đẳng thức xảy ra khi $a=b=c>0$ 

em thu lam cach khac anh xem nhe

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

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

$\dpi{150} \small $\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 \: \: \: \: \: \: \: \: \: \:$$