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

Question

Show that if $a$, $b$, $c$ and $d$ are non-negatives and $ab+bc+cd+da=1$ then: $$\sum\limits_{cyc} \frac {a^3} {b+c+d} \geq \frac {1} {3}$$

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)

Answer 1

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

Answer 2

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.

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@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.

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@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.

Answer 3

By Holder an AM-GM we obtain: $$\sum_{cyc}\frac{a^3}{b+c+d}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(b+c+d)}=\frac{1}{12}(a+b+c+d)^2\geq$$ $$\geq\frac{1}{12}\left(2\sqrt{(a+c)(b+d)}\right)^2=\frac{1}{3}(ab+bc+cd+da)=\frac{1}{3}.$$ Done!