RMO Inequalities - 1

By applying $AM-GM$ on $(1,a)$ , $(1,b)$ and $(1,c)$ : $\dfrac{1+a}2\geq\sqrt a;\quad\dfrac{1+b}2\geq\sqrt b;\quad \dfrac{1+c}2\geq\sqrt c$

Therefore, $\dfrac{(1+a)(1+b)(1+c)}{2×2×2}\geq\sqrt{abc}\\(1+a)(1+b)(1+c)\geq8\sqrt{abc}\\\implies1\geq\sqrt{abc}\\\boxed{abc\leq 1}$

Standard solution by Cauchy Schwarz:

$\begin{aligned} (1+a)(1+b)(1+c) &\geq (1+\sqrt[3]{abc})^3\\ 8 &\geq (1+\sqrt[3]{abc})^3\\ 1+\sqrt[3]{abc} &\leq 2\\ \sqrt[3]{abc} &\leq 1\\ abc &\leq 1\\ \end{aligned}$

Therefore, the maximum value of $abc=1$ .

As gernally8=2 2 2so a+1=2 a=1. Similarly b=1=a=c