[Toán 10] Bất đẳng thức

[nhockthongay_girlkute]

prove that if x,y,z >0 and [TEX]xyz=x+y+z+2[/TEX] then[TEX] 2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}) \leq x+y+z+6 [/TEX]

 

[nhockthongay_girlkute]

For non-negative real number a,b,c
Prove that :

[TEX]\sum \sqrt{ \frac{a(b+c)}{a^2+bc}} \geq 2 [/TEX]

solution

by AM-GM ,we have

[TEX]\frac{a^2+bc}{a(b+c)}+1 \geq 2\sqrt{\frac{a^2+bc}{a(b+c)} }[/TEX]

[TEX]<=> \frac{ (a+b)(a+c)}{a(b+c)} \geq 2\sqrt{\frac{a^2+bc}{a(b+c)}}[/TEX]

Reverse up and simple computation ,we want to prove

 

[vodichhocmai]

Cho a,b>0 thỏa mãn[TEX] (a+1)(b+1)=4[/TEX] CMR:
[TEX]a^2+b^2+6\geq4(a+b)[/TEX]

cho a,b>0 thỏa mãn (a+1)(b+1)=4 cmr:
[tex]a^2+b^2+6\geq4(a+b)[/tex]

[tex]LHS-RHS=^{\(gt\ \ a+b\ge 2\)}\(a+b\)\(a+b-2\)\ge 0\righ done!!\ \ [/tex]

 

[vodichhocmai]

prove that if x,y,z >0 and [TEX]xyz=x+y+z+2[/TEX] then


[TEX]4\( \frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)+6\le 2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}) \leq x+y+z+6 [/TEX]

 

[bigbang195]

Let tri ABC with area is S.prove that :

[TEX]\sum_{cyc}ab \ge 4\sqrt{3} S [/TEX]


Let ABC with area is S.prove that :

[TEX] \sum_{cyc}a^2 \ge 4\sqrt{3}S+\sum_{cyc}(a-b)^2[/TEX]

 

[namtuocvva18]

Cho a,b,c duong: Chung minh:
[TEX]\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\leq 3[/TEX].


Bai nay do doc nham de......
Sua nhu vay thi da tro thanh quyen thuoc .....
Nen em post bai khac vao day....


Cho a,b,c duong. Chung minh:
[TEX]\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a} \geq \frac{\sqrt{2}}{4}(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2})[/TEX].

 

[namtuocvva18]

1,Cho x,y,z duong. Tìm GTLN cua:
[TEX]P=\frac{xy}{x^2+xy+yz}+\frac{yz}{y^2+yz+zx}+\frac{zx}{z^2+zx+xy}[/TEX].

2,Cho a,b,c khong am thoả mãn [TEX]a+b+c=2[/TEX]. Tìm GTNN của:
[TEX]A=\frac{a}{3+b^2+c^2}+\frac{b}{3+c^2+a^2}+\frac{c}{3+a^2+b^2}[/TEX].

 

[namtuocvva18]

3,Cho a,b,c duong thoả mãn: [TEX]a^2+b^2+c^2=abc[/TEX]. Tìm GTLN của:
[TEX]B=\sqrt{\frac{a}{4a^2+7}}+\sqrt{\frac{b}{4b^2+7}}+\sqrt{\frac{c}{4c^2+7}}[/TEX].

4,Cho a,b,c la độ dài 3 cạnh tam giác co chu vi bàng 2010. Chung minh:
[TEX]\frac{ab}{1005-c}+\frac{bc}{1005-a}+\frac{ca}{1005-b}\geq 4020[/TEX].

 

[nhockthongay_girlkute]

let a,b,c be posiyive real numbers Prove that

[TEX]( \frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c})^2 \geq 4(ab+bc+ca)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})[/TEX]

 

[williamdunbar]

Cho a,b,c>0, [TEX]a^3+b^3+c^3=3[/TEX]. CMR:
[TEX]\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}\geq3[/TEX]

 

[nhockthongay_girlkute]

let a,b,c be the sides of triangle
Show that

[TEX]\sum \sqrt{a} ( \frac{1}{b+c-a}-\frac{1}{\sqrt{bc}}) \geq 0[/TEX]

 

[quyenuy0241]

[TEX]a,b,c>0 ,, a+b+c=3[/TEX]CMR:

[TEX]3abc+(a^2b^2+b^2c^2+c^2a^2) \ge abc(a^2+b^2+c^2)+3a^2b^2c^2 [/TEX]

 

[bigbang195]

[TEX]a,b,c>0 ,, a+b+c=3[/TEX]CMR:

[TEX]3abc+(a^2b^2+b^2c^2+c^2a^2) \ge abc(a^2+b^2+c^2)+3a^2b^2c^2 [/TEX]

[TEX]3r+(q^2-6r) \ge r(9-2q)+3r^2 \Leftrightarrow q^2-12r +2rq -3r^2 \ge 0[/TEX]

[TEX]3r^2+12r-q^2-2rq \le 0[/TEX]

delta should be positive, seems bad =, =

 

[vodichhocmai]

Let tri ABC with area is S.prove that :
[TEX]\sum_{cyc}ab \ge 4\sqrt{3} S [/TEX]
Let ABC with area is S.prove that :
[TEX] \sum_{cyc}a^2 \ge 4\sqrt{3}S+\sum_{cyc}(a-b)^2[/TEX]

[TEX]Newton:\ \ \(ab+bc+ca\)^2\ge 3abc\(a+b+c\) \ \ [/TEX]

 

[vodichhocmai]

Buồn chế chơi

Nếu như [TEX]a\ge b\ge c \ge 0 \ \ ;&\ \ \ \ a+b+c=1[/TEX] Chứng minh rằng khi đó ta có :
[TEX]a^2b+b^2c+c^2a \ge \sqrt{\frac{abc}{3}}[/TEX]

 

[vodichhocmai]

.........Gộp bài xoá ..............vodichhocmai..............................................

Cho a,b,c duong: Chung minh:
[TEX]\sqrt{\frac{2a}{a+b}}+\sqrt{\frac{2b}{b+c}}+\sqrt{\frac{2c}{c+a}}\leq 3[/TEX].

[TEX]xyz=1[/TEX] ta sẽ có BDT mạnh hơn

[TEX]\sqrt{\frac{2}{x^2+1}}\le \frac{3}{2}.\frac{(x+1\)}{x^2+x+1} =\frac{3}{2}\[1-\frac{x^2}{x^2+x+1} \][/TEX]


Mà cái thẳng cu cuối cùng thì rất wen thuộc bằng [TEX]Cauchy-Schwarz\ \ \ \ x=\frac{a^2}{bc}[/TEX]

 

[tell_me_goobye]

Find the min of the function
[TEX]f(x,y,z) =\frac{(1+2x)(3y+4x)(4y+3z)(2z+1)}{xyz}[/TEX] when x,y,z >0

 

[bigbang195]

For non-negative real number a,b,c
Prove that :

[TEX]\sum \sqrt{ \frac{a(b+c)}{a^2+bc}} \geq 2 [/TEX]

I'm find a solution very nice :

wlog : [TEX]a \geq b \geq c [/TEX]. then:

we have two easy inequalities: [TEX]\frac{a(b+c)}{a^2+bc} \ge \frac{b}{a}[/TEX] and [TEX]\frac{c(a+b)}{c^2+ab} \ge \frac{c}{b}.[/TEX]

hence :

[TEX]VT \ge \sqrt{\frac{b}{a}}+\sqrt{\frac{c}{b}}+\sqrt{\frac{b(a+c)}{b^2+ac}} \ge \sqrt{\frac{b}{a}+\frac{c}{b}}+\sqrt{\frac{b(a+c)}{b^2+ac}}=\sqrt{\frac{b^2+ac}{ab}}+\sqrt{\frac{b(a+c)}{b^2+ac}} \ge \sqrt{\frac{b^2+ac}{b(a+c)}}+\sqrt{\frac{b(a+c)}{b^2+ac}} \ge 2[/TEX]

Enjoy!!

 

[duynhan1]

Cho x,y,z dương thỏa :

[TEX]\huge x^{2009} + y^{2010} + z^{2011} \le x^{2008} + y^{2009} + z^{2010} [/TEX]

Chứng minh : [TEX]x+y+z \le 3 [/TEX]

 

[bigbang195]

tìm m để với mọi [TEX]x \in \mathbb{R}[/TEX] thì :

[TEX] f(x)=(m+1)x^2-2mx+2m \ge 0[/TEX]