The perimeter of $\triangle ABC$ is $2$ and it's sides are $BC=a, CA=b,AB=c$. Prove that $$abc+\frac{1}{27}\ge ab+bc+ca-1\ge abc. $$
Problem
Source: NMTC Junior P7
Tags: inequalities, AM-GM
Purple_Planet
02.11.2019 15:11
Notice that the inequality is equivalent to $$\frac{1}{27}\ge (1-a)(1-b)(1-c)\ge 0$$By triangle inequality $a+b\ge c\iff 2\ge 2c\iff 1-c\ge 0$. Similarly, $1-a\ge 0,1-b\ge 0$. So, $$(1-a)(1-b)(1-c)\ge 0$$By $AM-GM$
$$(1-a)(1-b)(1-c)\le \left(\frac{(1-a)+(1-b)+(1-c)}{3}\right)^3=\frac{1}{27}.$$
Math-wiz
02.11.2019 15:30
I usedthe substitution $x=s-a$, $y=s-b$, $z=s-c$, and then realised $s=1$, which made the question even easier
Tearfulstage
02.11.2019 16:36
I substituted as used schurs
sqing
05.11.2019 04:15
Purple_Planet wrote: The perimeter of $\triangle ABC$ is $2$ and it's sides are $BC=a, CA=b,AB=c$. Prove that $$abc+\frac{1}{27}\ge ab+bc+ca-1\ge abc. $$ NMTC 2019
Attachments:
Geronimo_1501
25.02.2020 20:22
Nothing. Scroll on..
WolfusA
25.02.2020 21:21
You found the maximum of LHS. It’s wrong.
Geronimo_1501
26.02.2020 09:13
Why do I do this all the time?