# Solution of problem T10/363 in M&Y $\left(a-\dfrac{1}{b}\right)\left(b-\dfrac{1}{c}\right)\left(c-\dfrac{1}{a}\right)\geq \left(a-\dfrac{1}{a}\right)\left(b-\dfrac{1}{b}\right)\left(c-\dfrac{1}{c}\right)\\ \forall a,b,c\in [1,+\infty).$

Solution of a my student.

Setting $a=x+1,b=y+1,c=z+1$, here $x,y,z$ are nonnegative real numbers, then we need prove $\prod [(x+1)(y+1)-1]\geq \prod [(x+1)^2-1]$ $\Leftrightarrow \sum_{\text{sym}}x^2y+\sum_{\text{cyclic}}x^2y^2\geq \sum_{\text{cyclic}}x^2yz+\sum_{\text{sym}}xyz$. This inequality is true, because by AM-GM we have $\sum_{\text{sym}}x^2y\geq \sum_{\text{sym}}xyz$ and $\sum_{\text{cyclic}}x^2y^2\geq \sum_{\text{cyclic}}x^2yz$.

## 3 thoughts on “Solution of problem T10/363 in M&Y”

1. psquang says:

Bat dang thuc tuong voi $\sum (a^2-1)(b-c)^2\ge 0$

2. psquang says:

Con co’ 1 cach’ khac’ nu~a tuy hoi dai do la: $(a-\frac{1}{b})(b-\frac{1}{a})\ge (a-\frac{1}{a})(b-\frac{1}{b})$

va chu’ y’ rang: $(a-\frac{1}{b})(b-\frac{1}{c})(c-\frac{1}{a})=(b-\frac{1}{a})(a-\frac{1}{c})(c-\frac{1}{b})$

3. NguyenDungTN says:

Quy dong len va dung BDT: $(ab-1)^2\ge (a^2-1)(b^2-1)$ (tuong duong voi $(a-b)^2 \ge 0$)
Tuong tu voi $b$ va $c$, nhan lai ta co DPCM.