[Toán 10] chứng minh BDT

B

bboy114crew

mình góp vui!
Bài 1: .Với [tex]a,b,c[/tex] la các số hữu tỉ. CMR:
[tex] (1+\frac{b-c}{a})^{a}+(1+\frac{c-a}{b})^{b}+(1+\frac{a-b}{c})^{c} \leq 1[/tex]
Bài 2: CMR:
[tex]2(\sqrt{a^{2}-ab+b^{2}}+\sqrt{b^{2}-bc+c^{2}}+\sqrt{c^{2}-ca+a^{2}})[/tex] [tex]\geq \sqrt{a^{2}+ab+b^{2}}+\sqrt{b^{2}+bc+c^{2}}+\sqrt{c^{2}+ca+a^{2}}[/tex]
Bài 3:: Cho [tex]a,b,c >0[/tex].CMR: [tex]\frac{2a^{3}}{a^{6}+bc}+\frac{2b^{3}}{b^{6}+ac}+\frac{2c^{3}}{c^{6}+ab}\leq \frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}[/tex]

Bài 4: Cho [tex]a,b,c >0[/tex]. Tìm GTNN của :
[tex]S=\sqrt{\frac{a^{3}}{a^{3}+(b+c)^{3}}}+\sqrt{\frac{b^{3}}{b^{3}+(a+c)^{3}}}+\sqrt{\frac{c^{3}}{c^{3}+(a+b)^{3}}}[/tex]
 
B

bboy114crew

mấy bài nữa !
BÀI 1:[tex]a,b,c > 0[/tex] CMR [tex]\sum \frac{b^2c}{a(a^2c+b^2c+1)} \geq \frac{ab+bc+ac}{1+2abc}[/tex]

BÀI 2:[tex]a,b,c > 0[/tex] CMR
[tex]\frac{a^4b}{(a^2+b)(b^2+c)} +\frac{b^4a}{(ab+c)(a+bc)}+\frac{c^3ab}{(1+c)(1+a)} \geq \frac{ab(a+b+c)}{(1+a)(1+b) [/tex]
 
You must log in or register to reply here.
Top Bottom