0

This is an exceptionally strange question. I have tried with all the methods I know. There is an equality which comes pretty close $(a+b)(b+c)(c+a)>8abc$. Please help me!!

Handuka
  • 1

1 Answers1

3

$a,b,c$ are positive numbers? If not, this inequality is not true, for example

$$a=b=-1,c=3.$$

So in the following we assume that $a,b,c>0$,and we can now use this inequality: $$(a+b)(b+c)(a+c)\ge\dfrac{8}{9}(a+b+c)(ab+bc+ac)\ge\dfrac{8}{9}(a+b+c)\cdot 3\sqrt[3]{a^2b^2c^2}\ge\dfrac{8}{3}(a+b+c)abc=8$$ because: $$3=(a+b+c)abc\ge 3(abc)^{\frac{4}{3}}\Longrightarrow abc\le 1$$ so: $$\sqrt[3]{a^2b^2c^2}\ge abc$$

math110
  • 93,304