giúp em nhé

Ta có: \begin{aligned} \frac{2023}{b+c}+\frac{2023}{c+a}-\frac{2023}{a+b+c} &=\frac{2023(a+b+c)}{(b+c)(c+a)}-\frac{2023}{a+b+c} \\ &=\frac{2023[(a+b+c)^2-(b+c)(c+a)]}{(b+c)(c+a)(a+b+c)} \\ &=\frac{2023[(a-b)^2+(a-c)(b-c)]}{(b+c)(c+a)(a+b+c)} \\ &>0 \end{aligned} Vì $a,b,c>0$ nên $(a-b)^2\geq 0$ và $(a-c)(b-c)\geq 0$, do đó $\frac{2023}{b+c}+\frac{2023}{c+a}>\frac{2023}{a+b+c}$.