27401

27401 (Radu Pop)

Fie $n\in \mathbb {N}$. Să se arate că

$$(n+1)ab(a+b)-(4n^{3}+13n+10)ab+(n+2)^{3}(a+b)\geq (n+2)^{3},$$

oricare ar fi $a,b\in [1,\infty )$

Soluție:

Fie $x,y\in [0,\infty )$. Avem

$$(x+1)(y+1)(x+y+n^{2}+n)=(n+1){\biggl (}\underbrace {{\frac {x}{n+1}}+{\frac {x}{n+1}}+\ldots +{\frac {x}{n+1}}} _{(n+1){\text{ ori}}}+1{\biggr )}\cdot$$

$$\cdot {\biggl (}\underbrace {{\frac {y}{n+1}}+{\frac {y}{n+1}}+\cdots +{\frac {y}{n+1}}} _{(n+1){\text{ ori}}}+1{\biggr )}\cdot {\biggl (}{\frac {x}{n+1}}+{\frac {y}{n+1}}+\underbrace {1+1+\cdots +1} _{n{\text{ ori}}}{\biggr )}\geq$$

$$\geq (n+1){\sqrt[{n+2}]{\frac {x^{n+1}}{(n+1)^{n+1}}}}\cdot {\sqrt[{n+2}]{\frac {y^{n+1}}{(n+1)^{n+1}}}}\cdot {\sqrt[{n+2}]{\frac {xy}{(n+1)^{2}}}}\cdot (n+2)^{3}=$$

$$={\frac {xy}{n+1}}(n+2)^{3}.$$

Rezultă că

$$(n+1)(x+1)(y+1)(x+y+n^{2}+n)\geq (n+2)^{3}xy.$$

Fie $x=a-1\geq 0$ şi $y=b-1\geq 0$.

Obţinem

$$(n+1)ab(a+b+n^{2}+n-2)\geq (n+2)^{3}(a-1)(b-1),$$

de unde

$$(n+1)ab(a+b)+(n^{3}+2n^{2}-n-2)ab\geq (n+2)^{3}ab-(n+2)^{3}(a+b)+(n+2)^{3}$$

deci

$$(n+1)ab(a+b)-(4n^{2}+13n+10)ab+(n+2)^{3}(a+b)\geq (n+2)^{3}.$$