2 replies to this topic

[Nguyen Thanh Toan]

1) Cho $a+b>1. CM: a^{4}+ b^{4}> \dfrac{1}{8}$
2) Cho $a+b=1. CM: (1+ \dfrac{1}{a}).(1+ \dfrac{1}{b})>9$

Edited by inhtoan, 23-06-2009 - 10:59.

[huyetdao_tama]

1) Cho $a+b>1. CM: a^{4}+ b^{4}> \dfrac{1}{8}$
2) Cho $a+b=1. CM: (1+ \dfrac{1}{a}).(1+ \dfrac{1}{b})>9$

Có:$a^4+b^4 \geq \dfrac{(a^2+b^2)^2}{2} \geq \dfrac{(a+b)^4}{8} > \dfrac{1}{8} $

Bài 2 a,b có dương không nhỉ.

[drnohad]

1) Cho $a+b>1. CM: a^{4}+ b^{4}> \dfrac{1}{8}$
2) Cho $a+b=1. CM: (1+ \dfrac{1}{a}).(1+ \dfrac{1}{b})>9$

Chắc là a,b đều dương hết
$VT=(1+1+\dfrac{a}{b})(1+1+\dfrac{b}{a}) \geq 3.3=9$(Cauchy 3 số)

Edited by drnohad, 23-06-2009 - 22:20.