# prove $\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$

Show that if $a,b,c,d \geq 0$ and $ab+bc+cd+da=1$ :$$\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$$ yet again it should be solved with Cauchy inequality.

thing i have done so far:

$(\sum\limits_{cyc} \frac {a^3} {b+c+d})\times(\sum\limits_{cyc} a(b+c+d)) \geq (\sum\limits_{cyc} a^2)^2$

so my problem is simplified to proving this:

$\frac {(\sum\limits_{cyc} a^2)^2} {(\sum\limits_{cyc} a(b+c+d))} \geq \frac {1} {3}$

$3 \times (\sum\limits_{cyc} a^2)^2 \geq \sum\limits_{cyc} a(b+c+d)$

$3 \times (a^2+b^2+c^2+d^2)^2 \geq 2(ab+ac+ad+bc+bd+cd)$

someone said to me if i play around with AM-GM it could be solved and i'm almost there

my idea is this right now:

prove $(a^2+b^2+c^2+d^2) \geq ab+bc+cd+da=1$ proved(with help of Jineon Baek hint)

prove $3(a^2+b^2+c^2+d^2) \geq 2(ab+ac+ad+bc+bd+cd)$ proved(with help of Jineon Baek hint)

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For your idea, just add up all inequalities of type $a^2+b^2 \geq 2ab$. –  Jineon Baek Jul 25 at 7:12
Jineon Baek’s hint works for both the inequalities you want to prove in your idea. –  Ewan Delanoy Jul 25 at 7:18
Jineon Baek, post your hint comment as an answer so i could accept it as an answer. –  user2838619 Jul 25 at 7:20

To finish your idea, just add up all inequalities of type $a^2+b^2 \geq 2ab$.

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This solution does not use Cauchy inequality, but I think it is interesting by its own.

Multiply the right-hand side by $ab+bc+cd+da$. Now the inequality is homogeneous so we can assume $a+b+c+d=1$ instead of $ab+bc+cd+da = 1$ by multiplying a common constant to all variables. Now we have to show this. $$\sum_{cyc} \frac{a^3}{1-a} \geq \frac{ab+bc+cd+da}{3}$$ By Cauchy or rearrangement inequality or whatever, $\sum_{cyc} a^2 \geq \sum_{cyc} ab$. So we only need to prove this. $$\sum_{cyc} \left( \frac{a^3}{1-a} - \frac{a^2}{3} \right) \geq 0$$ The function $f(x) = \frac{x^3}{1-x} - \frac{x^2}{3}$ is convex on $[0, 1]$. We just assumed $a+b+c+d=1$, so Jensen's inequality proves this.

@Macavity noticed that the function is not convex on whole interval, so this solution is not correct :( Since the interval where the function is convex is quite large, maybe we can divide the case by whether all variables are large enough or not. But that solution would be a mass.

@Macavity just gave a full solution for this. Instead of showing that $f$ is convex, it is enough to show that the tangent line of $f(x)$ at $x=1/4$ is less or equal to $f$. It seems to be a decent technique for proving competition-style inequalities.

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Close, but not quite done. $f$ is concave on $[0, \approx 0.0914]$. –  Macavity Jul 25 at 6:46
OTOH, you could show $$f(x) = \frac{x^3}{1-x} - \frac{x^2}{3}+\frac19(\frac14-x) \ge 0 \quad \forall x \in [0, 1]$$ Then what you need to prove is simply $f(a)+f(b)+f(c)+f(d) \ge 0$ –  Macavity Jul 25 at 6:50
@Macavity You're right. $f$ is not convex. Your second comment seems to work. –  Jineon Baek Jul 25 at 6:54
@Macavity $f(x)=\frac{(1-4x)^2(1+3x)}{36(1-x)}$ –  Ewan Delanoy Jul 25 at 7:16