If $a,b,c>0$ then $$\frac{1}{abc}+1\ge3\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{a+b+c}\right)$$
Problem
Source: Romanian National Olympiad 2019 - Grade 10
Tags: inequalities, romania
25.04.2019 16:18
ant_ wrote: If $a,b,c>0$ then $$\frac{1}{abc}+1\ge3\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{a+b+c}\right)$$ Let $abc=t^3$. Using AM-GM inequality, we get $$3\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{a+b+c}\right) \leq \frac{1}{t}+\frac{1}{t^2}=\frac{(t - 1)^2 (t + 1)}{t^3}+1+\frac{1}{t^3} \leq 1+\frac{1}{t^3}=1+\frac{1}{abc}$$
26.04.2019 05:07
ant_ wrote: If $a,b,c>0$ then $$\frac{1}{abc}+1\ge3\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{a+b+c}\right)$$ Proof of Zhangyunhua: By AM-GM, $$3\left(1+\frac{1}{abc}\right) =\left(\frac{1}{abc}+\frac{1}{abc}+1\right) +\left(\frac{1}{abc}+1+1\right) $$$$\geq \frac{3}{\sqrt[3]{a^2b^2c^2}}+\frac{3}{\sqrt[3]{abc}}\geq 9\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{a+b+c}\right).$$
26.04.2019 05:14
bel.jad5 wrote: \[=\frac{(t - 1)^2 (t + 1)}{t^3}+1+\frac{1}{t^3} \leq 1+\frac{1}{t^3}\]
12.05.2019 05:05
ant_ wrote: If $a,b,c>0$ then $$\frac{1}{abc}+1\ge3\left(\frac{1}{a^2+b^2+c^2}+\frac{1}{a+b+c}\right)$$
16.04.2020 13:45
nice solution for easy problem