Question

[Phụng Nguyễn Thị]

Chứng minh các bất đẳng thức :

a / \(\dfrac{a}{b}+\dfrac{b}{a}>=2;\forall a,b>0\)

b / \(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}>=3;\forall a,b,c>0\)

c / \(\left(a+b\right)\left(b+c\right)+\left(c+a\right)>=8abc;\forall a,b,c>=0\)

d / \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)>=9,\forall a,b,c>0\)

e / \(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)+\left(1+\dfrac{c}{a}\right)>=8,\forall a,b,c>0\)

f / \(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)>=4,\forall a,b,>0\)

HELP ME !!!!!!

Answer 1

[Ngọc Hồng]

a) Áp dụng BĐT AM - GM:

\(\dfrac{a}{b}+\dfrac{b}{a}\) >= 2\(\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}\) =2

Dấu '=' xảy ra <=> a=b=1

Answer 2

[Ngọc Hồng]

b) Cũng áp dụng BĐT AM- GM nhưng cho 3 số