\(\frac{\Rightarrow\left(m+n\right)\left(m^2+n^2\right)}{4}< =\frac{m^3+n^3}{2}\Rightarrow2\left(m+n\right)\left(m^2+n^2\right)< =4\left(m^3+n^3\right)\)
\(\Rightarrow2\left(m^3+n^3+m^2n+mn^2\right)< =4\left(m^3+n^3\right)\Rightarrow2\left(m^3+n^3\right)+2\left(m^2n+mn^2\right)< =\)
\(2\left(m^3+n^3\right)+2\left(m^3+n^3\right)\Rightarrow2\left(m^2n+mn^2\right)< =2\left(m^3+n^3\right)\)
\(\Rightarrow2\left(m^2n+mn^2\right)-2\left(m^3+n^3\right)=2\left(m^2n+mn^2-m^3-n^3\right)< =0\)
\(\Rightarrow2\left(\left(m^2n-m^3\right)+\left(mn^2-n^3\right)\right)=2\left(m^2\left(n-m\right)+n^2\left(m-n\right)\right)\)
\(=2\left(m^2\left(n-m\right)-n^2\left(n-m\right)\right)=2\left(m^2-n^2\right)\left(n-m\right)=2\left(m+n\right)\left(m-n\right)\left(n-m\right)\)
\(=-2\left(m+n\right)\left(m-n\right)\left(m-n\right)=-2\left(m+n\right)\left(m-n\right)^2< =0\)
vì \(-2< 0;m+n>0;\left(m-n\right)^2>=0\Rightarrow-2\left(m+n\right)\left(m-n\right)< =0\)luôn đúng
\(\Rightarrow\frac{m+n}{2}\cdot\frac{m^2+n^2}{2}< =\frac{m^3+n^3}{2}\)luôn đúng (đpcm)
dấu = xảy ra khi m=n
