a) Có: \(\left(a-1\right)^2\ge0,\forall a\)
\(\Leftrightarrow a^2-2a+1\ge0\)
\(\Leftrightarrow a^2+2a+1\ge4a\)
\(\Leftrightarrow\left(a+1\right)^2\ge4a\)
=>đpcm
b) Áp dụng bđt trên ta có:
\(\left(a+1\right)^2\ge4a\) (1)
\(\left(b+1\right)^2\ge4b\) (2)
\(\left(c+1\right)^2\ge4c\) (3)
Nhân vế vs vế (1) ; (2);(3) ta đc:
\(\left(a+1\right)^2\left(b+1\right)^2\left(c+1\right)^2\ge4a\cdot4b\cdot4c=64abc=64\)
\(\Rightarrow\left(a+1\right)\left(b+1\right)\left(c+1\right)\ge8\)
a) Theo Caucy thì: a2+b2>= 2ab.
=>(a+1)2=a2+1+2a>=4a
b) Theo Cauchy thì : a+b>=2\(\sqrt{ab}\)
b) Theo Cauchy thì a+b>=2\(\sqrt{ab}\)
=>a+1>=2\(\sqrt{a}\)
b+1>=2\(\sqrt{b}\)
c+1>=2\(\sqrt{c}\)
=> (a+1)(b+1)(c+1)>=8\(\sqrt{ }abc\)
=> (a+1)(b+1)(c+1)>=8
