2
$\begingroup$

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??

|cite|improve this question
$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
6
$\begingroup$

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 $$a^3+b^3=(a+b)(a^2-ab+b^2)\geq ab(a+b)$$

|cite|improve this answer
$\endgroup$
  • 1
    $\begingroup$ beautiful!!!!!! $\endgroup$ – jack Apr 13 '15 at 18:02
  • 1
    $\begingroup$ 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
1
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.