dungmathpro

Ta có:
$a+b+c=2007\Rightarrow \frac{1}{a+b+c}=\frac{1}{2007}\Rightarrow \frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
$\frac{1}{a+b+c}=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\Rightarrow \frac{1}{a+b+c}-\frac{1}{a}=\frac{1}{b}+\frac{1}{c}\Rightarrow \frac{-(b+c)}{a(a+b+c)}=\frac{b+c}{bc}\Rightarrow (b+c)\left [ \frac{1}{a(a+b+c)}+\frac{1}{bc} \right ]=0\Leftrightarrow (b+c)\left [ \frac{a^2+ab+ac+bc}{abc(a+b+c)} \right ]=0\Leftrightarrow (a+b)(b+c)(a+c)\left [ \frac{1}{abc(a+b+c)} \right]\Rightarrow (a+b)(b+c)(a+c)=0\Rightarrow \begin{bmatrix} a+b=0\\b+c=0 \\ c+a=0 \end{bmatrix}$
mà a+b+c=2007 suy ra một trong ba số bằng 2007
l

tìm nghiệm nguyên của phương trình sau:
$x^2y^2-x^2-3y^2+2x-1=0$

$\frac{a}{1+b^2c}= a-\frac{ab^2c}{1+b^2c}\geq a-\frac{ab^2c}{2b\sqrt{c}}=a-\frac{ab\sqrt{c}}{2}=a-\frac{b\sqrt{a.ac}}{2}\geq a-\frac{b(a+ac)}{4}$
tương tự:
$P\geq (a+b+c+d)-\frac{1}{4}(ab+bc+cd+ad+abc+bcd+cda+dab)$
lại có:
$ab+bc+cd+ad\leq \frac{1}{4}(a+b+c+d)^2$
$abc+bcd+cda+dab\leq \frac{1}{16}(a+b+c+d)^3$
$\Rightarrow P\geq 4-\frac{1}{4}(4+4)=2$

Ta có
$6a^3+\frac{3}{2}a^2+\frac{3}{8}a\geq 3.\frac{3}{2}a^2=\frac{9}{2}a^2\Rightarrow 6a^3\geq 3a^2-\frac{3}{8}a$
tương tự:
$\Rightarrow 6(a^3+b^3+c^3+d^3)\geq 3(a^2+b^2+c^2+d^2)-\frac{3}{8}(a+b+c+d)= (a^2+b^2+c^2+d^2)+2(a^2+b^2+c^2+d^2)-\frac{3}{8}\geq (a^2+b^2+c^2+d^2)+2.\frac{1}{4}(a+b+c+d)^2-\frac{3}{8}=(a^2+b^2+c^2+d^2)+\frac{1}{2}-\frac{3}{8}=\frac{1}{8}\Rightarrow đpcm$

Ta có
$\frac{1}{\sqrt{a^2+ab+b^2}}+\frac{1}{\sqrt{b^2+bc+c^2}}+\frac{1}{\sqrt{a^2+ac+c^2}}\geq \frac{9}{\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{a^2+ac+c^2}}$
Để ý:
${\sqrt{a^2+ab+b^2}+\sqrt{b^2+bc+c^2}+\sqrt{a^2+ac+c^2}}\leq \frac{a^2+ab+b^2+b^2+bc+c^2+a^2+ac+c^2+3}{2}\leq \frac{3(ab+bc+ac)+ab+ac+bc+2(a^2+b^2+c^2)}{2}=\frac{2(a+b+c)^2}{2}=(a+b+c)^2$
Vậy
$\frac{1}{\sqrt{a^2+ab+b^2}}+\frac{1}{\sqrt{b^2+bc+c^2}}+\frac{1}{\sqrt{a^2+ac+c^2}}\geq \frac{9}{(a+b+c)^2}\Rightarrow đpcm$

minh chem cau hai nha!!!!!!
$\frac{1}{a+2}+\frac{2008}{2009+b}\leq \frac{c+1}{2008+c}$
dat x=a+1;y=b+1;z=c+1
BDT tuong duong
$\frac{1}{x+1}+\frac{2008}{2008+y}\leq \frac{z}{2007+z}\Leftrightarrow \frac{1}{x+1}+\frac{2008}{2008+y}+\frac{2007}{2007+z}\leq \bigl(\begin{smallmatrix} 1 \end{smallmatrix}\bigr)$
tu 1 theo BDT co si ta co
$\frac{x}{x+1}= 1-\frac{1}{x+1}=\frac{2008}{2008+y}+\frac{2007}{2007+z}\geq 2\sqrt{\frac{2008.2007}{(2008+y)(2007+z)}}$
tuong tu
$\frac{y}{y+2008}=\geq 2\sqrt{\frac{2007}{(x+1)(2007+z)}}$
$\frac{z}{2007+z}\geq 2\sqrt{\frac{2008}{(x+1)(y+2008)}}$
Nhan VTV
$xyz\geq 8.2007.2008=32240448$

Bai 2
dat: $\frac{2m+1}{m^2+1}=y\Leftrightarrow ym^2-2m+y-1=0$
xet y=0$\fra\Rightarrow m=\frac{-1}{2}$
xet y#0 ta co:
pt $ym^2-2m+y-1$ co$\Delta =-4y^2+4y+4\Rightarrow \frac{1-\sqrt{5}}{2}\leq y\leq \frac{1+\sqrt{5}}{2}$
vay GTLN cua bieu thuc la:
$Max=\frac{1+\sqrt{5}}{2}$

$5(a+b+c)+\frac{3}{abc}\geq 5(a+b+c)+\frac{a^2+b^2+c^2}{(\frac{a+b+c}{3})^3}\geq 5(a+b+c)+\frac{\frac{(a+b+c)^2}{3}}{(\frac{a+b+c}{3})^3}= 5(a+b+c)+\frac{9}{a+b+c}= \frac{45}{a+b+c}+5(a+b+c)-\frac{36}{a+b+c}\geq 30-\frac{36}{a+b+c}$
de y
$a+b+c\leq \sqrt{3(a^2+b^2+c^2)}=3\Rightarrow \frac{36}{a+b+c}\geq 12\Rightarrow -\frac{36}{a+b+c}\leq -12$
suy ra $P\geq 30-12=18\Rightarrow dpcm$

Giai:
ta co
$\frac{\sqrt{b^2+2a^2}}{ab}=\sqrt{}\frac{{b^2+2a^2}}{a^2b^2}=\sqrt{\frac{1}{a^2}+\frac{2}{b^2}}$
tuong tu
dat $x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\Rightarrow x+y+z=1$
ta co
$3(x^2+2y^2)\geq (x+2y)^2\Rightarrow \sqrt{x^2+2y^2}\geq \frac{1}{\sqrt{3}}(x+2y)$
tuong tu cong lai ta dc
P$P\geq \frac{1}{\sqrt{3}}(3x+3y+3z)=\sqrt{3}$

Bài 1:
$a^5+b^5=(a^3+b^3)(a^2+b^2)-a^2b^2(a+b)\geq ab(a+b)2ab-a^2b^2(a+b)= a^2b^2(a+b) \Rightarrow a^5+b^5+ab\geq ab(a^2b+ab^2+abc)\Rightarrow \frac{ab}{a^5+b^5+ab}\leq \frac{1}{ab(a+b+c)}$
tương tự
$\Rightarrow A\leq \frac{1}{ab(a+b+c)}+\frac{1}{bc(a+b+c)}+\frac{1}{ac(a+b+c)}=\frac{a+b+c}{abc(a+b+c)}=1\Rightarrow đpcm$

1.Cho x,y,z là số thực dương và $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq 2$
Tìm giá trị lớn nhất của biểu thức P=(x-1)(y-1)(z-1)
(Trích câu 2 đề thi năm 2012 tỉnh quảng trị lớp 11)
2.cho a,b,c>0 tìm giá trị nhỏ nhất của biểu thức
$\frac{a}{b+c-a}+\frac{b}{c+a-b}+\frac{c}{b+a-c}$
(trích câu 4 đề thi hsg lớp 11 tỉnh quảng trị năm 2011)

Dat Bieu thuc tren la A
$$\frac{ab}{\sqrt{ab+2c}}+\frac{bc}{\sqrt{bc+2a}}+\frac{ac}{\sqrt{ac+2b}}=\frac{ab}{\sqrt{(a+c)(b+c)}}+\frac{bc}{\sqrt{(a+b)(a+c)}}+\frac{ab}{\sqrt{(a+b)(b+c)}}$$
Ap dung Cauchy ta co
$\frac{ab}{\sqrt{(a+c)(b+c)}}\leq \frac{a^{2}}{2(a+c)}+\frac{b^{2}}{2(b+c)}$
tuong tu
$A\leq \frac{a^2}{2(a+c)}+\frac{b^2}{2(b+c)}+\frac{b^2}{2(a+c)}+\frac{c^2}{2(a+b)}+\frac{a^2}{2(a+b)}+\frac{c^2}{2(b+c)}$
Ap dung Bunhiacopxki ta co:
$A\leq 2\frac{(a+b+c)^2}{4(a+b+c)}= 1$
Vay max A=1

$(x_{1}-x_{2})^2=(x_{1}+x_{2})^2-4x_{1}.x_{2}\Rightarrow x_{1}-x_{2}=\pm \sqrt{(x_{1}+x_{2})^2-4x_{1}.x_{2}}$
Ap dung he thuc Viet ta co dpcm

BĐT 19:
Với a,b,c là 3 số thực dương ta có
$\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}\geq \dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}$


Chứng minh:
Áp dụng BĐT Cauchy-Schwarz ta có
$3(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2})\geq (\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a})^2$
Áp dụng BĐT AM-GM ta được $(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2})\geq 3$
$3(\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2})^2\geq 3(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a})^2$
Từ đây ta có đpcm

Cach khac:
$\frac{a^2}{b^2}+1\geq 2\frac{a}{b} \Rightarrow \frac{a^2}{b^2}+\frac{b^2}{c^2}+\frac{c^2}{a^2}\geq 2.\left ( \frac{a}{b}+\frac{b}{c}+\frac{c}{a} \right )-3$
lại có: $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq 3\Rightarrow dpcm$