Show that if $a,b,c$ are positive reals and $a+b+c=1$, then the following must hold:

$$\frac{2(a^3+b^3+c^3)}{abc}+3 \geq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}$$

What I have tried is using $abc \leq \frac{1}{27}$ $(a+b+c \geq 3\sqrt[3]{abc}) $ and multiplying everything by $abc$, but I don'think that's a good idea because $abc$ can be positive and negative. I have also tried substituting $a^3+b^3+c^3 \geq 3abc$, but that isn't strong enough. Any help/hints please??

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    $\begingroup$ This looks like an application of the AM-GM inequality $\endgroup$ – Surb Apr 13 '15 at 17:23
  • $\begingroup$ Probably, but I have no idea where/how. $\endgroup$ – jack Apr 13 '15 at 17:24
  • $\begingroup$ what kind of numbers are $a,b,c$? $\endgroup$ – Dr. Sonnhard Graubner Apr 13 '15 at 17:29
  • $\begingroup$ It is not true for $(a,b,c)=(1/2,-1,3/2)$, for example. $\endgroup$ – mathlove Apr 13 '15 at 17:30
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    $\begingroup$ Note that $a,b,c$ cannot be $0$ since that makes the expressions undefined. I think $a,b,c$ are positive reals. $\endgroup$ – Prasun Biswas Apr 13 '15 at 17:37

if $a,b,c>0$ then we get $$\frac{2(a^3+b^3+c^3)}{abc}+3\geq \frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c}$$ this is equivalent to $$\frac{2(a^3+b^3+c^3)}{abc}\geq \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}$$ and this $$2(a^3+b^3+c^3)\geq bc(b+c)+ac(a+c)+ab(a+b)$$ and now note that $$a^3+b^3=(a+b)(a^2-ab+b^2)\geq ab(a+b)$$

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    $\begingroup$ beautiful!!!!!! $\endgroup$ – jack Apr 13 '15 at 18:02
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    $\begingroup$ or rearrangement for last line $\endgroup$ – jack Apr 13 '15 at 18:11
  • $\begingroup$ the next problem please, i love inequalities! $\endgroup$ – Dr. Sonnhard Graubner Jan 25 '16 at 17:08

Firstly, multiply the right-hand side by $a+b+c$, so that both sides are now degree-zero. Then you want to prove $$2a^3+2b^3+2c^3\geq a^2b+ab^2+a^2c+ac^2+b^2c+bc^2$$

  • $\begingroup$ rearrangement inequality I guess $\endgroup$ – jack Apr 13 '15 at 18:06
  • $\begingroup$ @jack Or $a^3+b^3=(a+b)(a^2-ab+b^2)\ge ab(a+b)$ (which is simple to prove -- $a^2-ab+b^2\ge ab\iff (a-b)^2\ge 0$ with equality iff $a=b$). This is actually identical to Sonnhard's solution, except for this last claim. $\endgroup$ – user26486 Apr 13 '15 at 18:16
  • $\begingroup$ What last claim? He uses the same, no? $\endgroup$ – jack Apr 13 '15 at 18:21
  • $\begingroup$ @jack Michael doesn't prove it while Sonnhard does. $\endgroup$ – user26486 Apr 13 '15 at 18:28
  • $\begingroup$ you mean the opposite, Michael does prove it $\endgroup$ – jack Apr 13 '15 at 20:37

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