Question

[Dũng Quang]

Cho a,b,c >0. Chứng minh:

a) ab+a/b+b/a>= a+b+1

b) a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)>=6abc

c) (a+b)(b+c)(c+a)>=8abc

d) ab/c+bc/a+ac/b>=a+b+c

Answer 1

[Akai Haruma]

b/

Áp dụng BĐT Cô-si ta có:

$a^2(1+b^2)+b^2(1+c^2)+c^2(1+a^2)=a^2+b^2+c^2+a^2b^2+b^2c^2+c^2a^2$
$=(a^2+b^2c^2)+(b^2+c^2a^2)+(c^2+a^2b^2)$

$\geq 2\sqrt{a^2b^2c^2}+2\sqrt{b^2c^2a^2}+2\sqrt{c^2a^2b^2}$

$=2abc+2abc+2abc=6abc$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c=1$

Answer 2

[Akai Haruma]

c/

Áp dụng BĐT Cô-si:

$(a+b)(b+c)(c+a)\geq 2\sqrt{ab}.2\sqrt{bc}.2\sqrt{ca}=8abc$

Ta có đpcm

Dấu "=" xảy ra khi $a=b=c$.