5
$\begingroup$

How to prove this for positive real numbers? $$a+b+c\leq\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}$$

I tried AM-GM, CS inequality but all failed.

$\endgroup$
  • 1
    $\begingroup$ I hope it is given that $a,b,c \gt 0$. $\endgroup$ – Indrayudh Roy Feb 17 '14 at 14:42
  • $\begingroup$ Why somebody downvote? $\endgroup$ – zhangwfjh Feb 17 '14 at 14:44
  • $\begingroup$ Does $a$, $b$, $c$ are arbitrary real numbers? Or they are positive numbers? $\endgroup$ – Lion Feb 17 '14 at 14:50
  • $\begingroup$ @Lion Non zero reals. $\endgroup$ – zhangwfjh Feb 17 '14 at 14:52
  • $\begingroup$ Do you have try the method of extreme value of multivariable function? $\endgroup$ – Lion Feb 17 '14 at 14:54
11
$\begingroup$

Using Cauchy-Schwarz Inequality twice:

$a^4 + b^4 +c^4 \geq a^2b^2 +b^2c^2 +c^2a^2 \geq ab^2c +ba^2c +ac^2b = abc(a+b+c)$

$\endgroup$
  • $\begingroup$ Substitute $a,b,c$ in $a^2 + b^2 +c^2 \geq ab +bc +ca$ with $ab,bc,ca$ respectively. $\endgroup$ – r9m Feb 17 '14 at 14:49
  • $\begingroup$ If $abc<0$ then you get the reverse inequality, right? $\endgroup$ – user37238 Feb 17 '14 at 14:54
  • $\begingroup$ $a^2 + b^2 +c^2-ab-bc-ca = 1/2((a-b)^2 +(b-c)^2 +(c-a)^2) \geq 0$ so it holds for all $a,b,c$ $\endgroup$ – r9m Feb 17 '14 at 15:00
  • $\begingroup$ Please try $a=-2$, $b=1$ and $c=1$. $\endgroup$ – user37238 Feb 17 '14 at 15:02
  • $\begingroup$ @user129017, when you divide both sides by $abc$, then the inequality gets reversed when $abc \lt 0$. $\endgroup$ – Indrayudh Roy Feb 17 '14 at 15:03
7
$\begingroup$

I have come up with an answer with myself. Using CS inequality $$(a^4+b^4+c^4)(1+1+1)\geq(a^2+b^2+c^2)^2$$ $$(a^2+b^2+c^2)(1+1+1)\geq(a+b+c)^2$$ Hence we have $$a^4+b^4+c^4\geq\frac{(a+b+c)^4}{27}=(a+b+c)\left(\frac{a+b+c}{3}\right)^3\geq abc(a+b+c)$$

$\endgroup$
  • $\begingroup$ Your answer is invalid. $$\left(\frac{a+b+c}{3}\right)\ge abc$$ does not hold for all $a,b,c\in\mathbb R$. Mean inequalities are only used for positive numbers (except for $\frac{a+b}{2}\ge \sqrt{ab}$). If you still don't believe me, take, e.g., $a=-1$, $b=-1$ and $c=1$. $\endgroup$ – user26486 Feb 22 '14 at 0:51
7
$\begingroup$

Here other two answers used Cauchy-Scwartz Inequality. I am giving a simple $AM\ge GM$ inequality proof.

You asked, $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c\\\implies a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$$

Now, from, $AM\ge GM$, we have $$\frac {a^4+ a^4+b^4+c^4}4\ge \left(a^4\cdot a^4\cdot b^4\cdot c^4\right)^{1/4}=a^2bc\tag 1$$

Similarly, $$\frac {a^4+ b^4+b^4+c^4}4\ge \left(a^4\cdot b^4\cdot b^4\cdot c^4\right)^{1/4}=ab^2c\tag 2$$ and also, $$\frac {a^4+ b^4+c^4+c^4}4\ge \left(a^4\cdot b^4\cdot c^4\cdot c^4\right)^{1/4}=abc^2\tag 3$$

Now, summing up $(1),(2),(3)$, we have, $a^4+b^4+c^4\ge a^2bc+b^2ca+c^2ab$, that is $$\frac{a^3}{bc}+\frac{b^3}{ca}+\frac{c^3}{ab}\ge a+b+c$$

$\endgroup$
4
$\begingroup$

By Holder $$\sum_{cyc}\frac{a^3}{bc}\geq\frac{(a+b+c)^3}{3(ab+ac+bc)}=\frac{(a+b+c)\cdot(a+b+c)^2}{3(ab+ac+bc)}\geq a+b+c$$

$\endgroup$

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.