$\\$ $\sqrt{2024a+bc}=\sqrt{(a+b+c)a+bc}=\sqrt{(a^{2}+ab)+(bc+ca)}=\sqrt{a(a+b)+(a+b)c}=\sqrt{(a+b)(a+c)}$ $\\$ $Áp$ $dụng$ $bất$ $đẳng$ $thức$ $Buniacopxki$ $:$ $\\$ =>$(a+b)(c+a) \ge \sqrt{ac}+\sqrt{ab}$ $\\$ =>$\frac{a}{a+\sqrt{2024a+bc}} \le \frac{a}{a+\sqrt{ac}+\sqrt{ab}}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$ $\\$ $Chứng$ $minh$ $tương$ $tự$ $:$ $\\$ =>$\frac{b}{b+\sqrt{2024b+ac}} \le \frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$ $\\$ =>$\frac{c}{c+\sqrt{2024c+ab}} \le \frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}$ $\\$ =>$\frac{a}{a+\sqrt{2024a+bc}} + \frac{b}{b+\sqrt{2024b+ac}} + \frac{c}{c+\sqrt{2024c+ab}} \le \frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}} +\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}} + \frac{\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1$ $\\$ =>$Điều$ $phải$ $chứng$ $minh$ $\\$ $Dấu$ $"="$ $khi$ $a=b=c=\frac{2024}{3}$Đáp án:

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