[Toán 10]Bdt

[bigbang195]

[TEX]\blue a,b,c>0[/TEX]

[TEX]\blue \sqrt[3]{\frac{(a+b)(b+c)(a+c)}{8}} \ge \sqrt{\frac{ab+bc+ac}{3}}[/TEX]

[TEX]\red (Carlson's \: \:inequality)[/TEX]

 

[bigbang195]

[TEX]\blue a,b,c>0[/TEX].

[TEX]\blue a^2+b^2+c^2+2abc+3 \ge (1+a)(1+b)(1+c)[/TEX]

 

[bigbang195]

[TEX]\blue a,b,c>0[/TEX]

[TEX]\blue \sum_{cyclic} \frac{a}{1+b} \ge \frac{3}{1+abc}[/TEX]

 

[bigbang195]

[TEX]\blue a,b,c \in R.[/TEX]

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

[TEX]\red Vasile \:\: Cirtoaje[/TEX]

 

[bigbang195]

[TEX]\blue a,b,c>0. [/TEX]

[TEX]\blue \sum_{cyclic} \sqrt{ab(a+b)} \ge \sqrt{4abc+(a+b)(a+c)(b+c)}[/TEX]

 

[bigbang195]

[TEX]\blue a,b,c>0[/TEX]

[TEX]\blue \sum_{cyclic} \frac{a}{1+b} \ge \frac{3(a+b+c)}{a+b+c+3}[/TEX]

 

[bigbang195]

[TEX]\blue a,b,c>0[/TEX]

[TEX]\blue \sum_{cyc} \sqrt{\frac{a^3}{a^3+(b+c)^3}} \ge 1[/TEX]

 

[king_math96]

[TEX]\blue a,b,c>0[/TEX]

[TEX]\blue \sum_{cyc} \sqrt{\frac{a^3}{a^3+(b+c)^3}} \ge 1[/TEX]

ta chỉ cần Cm bất đẳng thức:
[TEX]\sqrt{\frac{a^3}{a^3+(b+c)^3}} \geq \frac{a^2}{a^2+b^2+c^2}[/TEX]

 

[quyenuy0241]

[TEX]\blue a,b,c \ge 0[/TEX] thỏa mãn [TEX]\blue a+b+c \ge abc[/TEX] . CM

[TEX]\blue a^2+b^2+c^2 \ge \sqrt{3}abc[/TEX]

[TEX]\red (BMO \: \: 2001)[/TEX]

[tex](a^2+b^2+c^2)^2 \ge 3abc(a+b+c) \ge 3a^2b^2c^2 ,,\fbox{Done!!}[/tex]

 

[quyenuy0241]

cho a,b,c khong am

CMR:
[tex](ab+c^2)(bc+a^2)(ca+b^2) \ge abc(a+b)(b+c)(c+a)[/tex]

 

[quyenuy0241]

[TEX]\red Hong\: Kong \: 2000[/TEX]

[TEX]\blue abc=1[/TEX] và [TEX]\blue a,b,c>0[/TEX] . CM

[TEX]\blue \sum_{cyc} \frac{1+ab^2}{c^3} \ge \frac{18}{a^3+b^3+c^3}[/TEX]

[tex]VT= \sum{a^3b^3(1+ab^2)=\sum{a^3b^3+\sum{a^4b^5} [/tex]

[tex]\sum{a^3b^3 \ge 3a^2b^2c^2=3[/tex]

[tex]\sum{a^4b^5 \ge 3a^3b^3c^3=3 [/tex]

[tex]VT \ge 6 [/tex]

[tex]VP=\frac{18}{a^3+b^3+c^3} \le 6[/tex]

[tex]VT \ge 6 \ge Vp [/tex]

 

[quyenuy0241]

[TEX]\blue a,b,c \in R.[/TEX]

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

[TEX]\red Vasile \:\: Cirtoaje[/TEX]

[tex] (a^2+b^2+c^2)^2-3a^3b-b^3c-c^3a=\frac{1}{2}\sum(x^2-y^2-xy+2yz-zx)^2[/tex]

Hình như giống bài của anh vodichhocmai thì phải @-)

 

[quyenuy0241]

[TEX]\blue a,b,c>0[/TEX]

[TEX]\blue \sqrt[3]{\frac{(a+b)(b+c)(a+c)}{8}} \ge \sqrt{\frac{ab+bc+ac}{3}}[/TEX]

[TEX]\red (Carlson's \: \:inequality)[/TEX]

CHuẩn hóa : ab+bc+ac=3

Cần CM :

[tex](a+b)(b+c)(a+c) \ge 8[/tex]

[tex](a+b)(a+c)(b+c) \ge \frac{8}{9}(a+b+c)(ab+bc+ac) \ge 8 [/tex]

 

[huycuopbien123]

cho a,b,c khong am

CMR:
[tex](ab+c^2)(bc+a^2)(ca+b^2) \ge abc(a+b)(b+c)(c+a)[/tex]

[tex]holder\Rightarrow (1+\frac{a^2}{bc})^{\frac{2}{3}}(1+\frac{b^2}{ca})^{\frac{1}{3}}\geq 1+\frac{a}{c}[/tex]
Tương tự [tex](1+\frac{b^2}{ac})^{\frac{2}{3}}(1+\frac{c^2}{ba})^{\frac{1}{3}}\geq1+\frac{b}{a}[/tex]
[tex](1+\frac{c^2}{ab})^{\frac{2}{3}}(1+\frac{a^2}{bc})^{\frac{1}{3}}\geq1+\frac{c}{b}[/tex]
Nhân theo vế [tex]\Rightarrow (1+\frac{a^2}{bc})(1+\frac{b^2}{ac})(1+\frac{c^2}{ab})\geq (1+\frac{a}{c})(1+\frac{b}{a})(1+\frac{c}{b}) \Leftrightarrow dpcm[/tex]

 

[bigbang195]

cho a,b,c khong am

CMR:
[tex](ab+c^2)(bc+a^2)(ca+b^2) \ge abc(a+b)(b+c)(c+a)[/tex]

Ta chứng minh BDT mạnh hơn như sau:

[TEX]\blue \prod_{cyclic}(a^2+bc) \ge 4abc\sqrt[3]{\prod_{cyclic}(a^3+b^3)}[/TEX]

[TEX]\blue (a^2+bc)(b^2+ac)=ab(c^2+ab)+c(a^3+b^3) \ge 2\sqrt{abc(a^3+b^3)(c^2+ab)[/TEX]

[TEX]\blue \Rightarrow \left (\prod_{cyclic}(a^2+bc) \right)^2 \ge 8\sqrt{a^3b^3c^3\prod_{cyclic}(a^3+b^3)\prod_{cyclic}(a^2+bc)}[/TEX]

[TEX]\fbox{Done!!}[/TEX]
)

 

[bigbang195]

Cách khác ta chứng minh

[TEX]\blue (a^2+bc)(b^2+ac)-ab(a+c)(b+c) \ge 0[/TEX]

[TEX]\blue \Leftrightarrow c(a-b)^2(a+b) \ge 0[/TEX]

Bằng cách thiết lập 2 bất đẳng thức còn lại rồi nhân lại ta thu được điều phải chứng minh.

dấu bằng xảy ra khi và chỉ khi [TEX]\blue a=b=c[/TEX]

 

[bigbang195]

[TEX]\blue a,b,c[/TEX] là các số thực dương thoả mãn [TEX]\blue ab+bc+ac+abc=4[/TEX]. Chứng minh rằng:

[TEX]\fbox{\blue a+b+c \ge ab+bc+ac [/TEX]

[TEX]\fbox{\red Viet\: Nam\: 1996[/TEX]

 

[bigbang195]

Easy but Nice !!

[TEX]\fbox{\red Lasvia\: 2002[/TEX]

[TEX]\blue a,b,c,d [/TEX]là các số thực dương thoả mãn :

[TEX]\blue \sum_{a,b,c,d} \frac{1}{a^4+1}=1[/TEX].

Chứng minh:

[TEX]\fbox{ \blue abcd \ge 3[/TEX]

 

[bigbang195]

[TEX]\blue [Vasile\: Cirtoaje][/TEX]

Let a,b,c be positive real number .Prove that

[TEX]\blue \sum_{cyclic} \frac{a^2}{b^2+c^2} \ge \sum_{cyclic} \frac{a}{b+c}[/TEX]

 

[bigbang195]

Let [TEX]\blue x,y,z[/TEX] be positive real number such that [TEX]\blue xyz=x+y+z+2[/TEX].Prove that

[TEX]\blue (1)xy+yz+xz \ge 2(x+y+z)[/TEX]

[TEX]\blue (2)\sqrt{x}+\sqrt{y}+\sqrt{z} \le \frac{3}{2}\sqrt{xyz}[/TEX]