Question

cm bất đẳng thức:

a, a\(^2\)+ b\(^2\)+2 \(\ge\) 2.(a+b)

b, \(\frac{a+b}{2}\) . \(\frac{a^2+b^2}{2}\) \(\le\) \(\frac{a^3+b^3}{2}\) ( với a,b >0 )

Answer 1

\(a^2+b^2+2\ge2\left(a+b\right)\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2\ge0\) (đúng)

\("="\Leftrightarrow a=b=1\)

\(\frac{a+b}{2}.\frac{a^2+b^2}{2}\le\frac{a^3+b^3}{2}\Leftrightarrow\left(a+b\right)\left(a^2+b^2\right)\le2\left(a^3+b^3\right)\)

\(\Leftrightarrow a^2b+ab^2\le a^3+b^3\Leftrightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)(đúng)

\("="\Leftrightarrow a=b\)