[TEX]\Leftrightarrow \frac{(a^3+b^3)}{2}\leq \frac{(a^2+b^2)^3}{(a+b)^3}[/TEX]
Thử xé tung cái này ra xem sao.Trước hết,ta chứng minh:
[TEX](a^3+b^3)(a^3+b^3+3a^2b+3ab^2)\leq 2(a^6+b^6+3a^4b^2+3a^2b^4)[/TEX]
[TEX]\Leftrightarrow a^6+a^3b^3+3a^5b+3a^4b^2+a^3b^3+b^6+3a^2b^4+3ab^5 \leq 2(a^6+b^6+3a^4b^2+3a^2b^4)[/TEX]
[TEX]\Leftrightarrow a^6+b^6-2a^3b^3+3ab(a^3b+ab^3-a^4-b^4)\geq 0[/TEX]
[TEX]\Leftrightarrow (a^3-b^3)^2+3ab(a-b)^2(a^2+ab+b^2)\geq 0[/TEX](BĐT đúng)
[TEX]\Rightarrow (a^3+b^3)(a+b)^3\leq 2(a^2+b^2)^3[/TEX]
[TEX]\Rightarrow dpcm[/TEX]