Prove for a,b,c,d > 0 $\frac{a^3}{b+c+d} + \frac{b^3}{a+c+d} + \frac{c^3}{a+b+d} + \frac{d^3}{a+b+c} \ge \frac{1}{3}$ where $ab + bc + cd + da = 1$

Question

The task requires for Cauchy-Schwarz inequality to be used.

My attempt: Using Titu's Lemma (direct consequence of the inequality), I got:

$\frac{a^4}{a(b+c+d)} + \frac{b^4}{b(a+c+d)} + \frac{c^4}{c(a+b+d)} + \frac{d^4}{d(a+b+c)} \ge \frac{(a^2 + b^2 + c^2 + d^2)^2}{a(b+c+d) \ + \ b(a+c+d) \ + \ c(a+b+d) \ + \ d(a+b+c)}$

Using the condition that $ab + bc + cd + da = 1$, the RHS results in $\frac{(a^2+b^2+c^2+d^2)^2}{2+2(ac+bd)}$.

Nothing more seems to help. Opening up the numerator or using AM-GM on it doesn't work and I have no idea on how to turn this result in a $\frac{1}{3}$ fraction.

Answer 1

By Cauchy and AM-GM inequality, $$\sum\frac{a^3}{b+c+d}\ge\frac{\left(\sum a^2\right)^2}{2\sum ab}\ge\frac{\left(\sum a^2\right)^2}{3\sum a^2}=\frac{\sum a^2}3\ge\frac13.$$ The last "$\ge$" is equivalent to $\sum a^2\ge1$. By rearrangement inequality $$a^2+b^2+c^2+d^2\ge ab+bc+cd+da=1.$$We're done.

Answer 2

\begin{align*} \sum \frac{a^3}{b+c+d} &\geq \frac{4(a^3+b^3+c^3+d^3)}{3(a+b+c+d)}\\ &\geq \frac{(a+c)^3+(b+d)^3}{3(a+b+c+d)}\\ &=\frac{1}{3}((a+c)^2+(b+d)^2-(a+c)(b+d))\\ &\geq \frac{(a+c)(b+d)}{3}\\ &=\frac{1}{3} \end{align*}

Answer 3

Continuing from where you left off, note the condition is equivalent to $(a+c)(b+d)=1$. Then reduce the numerator using $$ a^2+b^2+c^2+d^2 = (a+c)^2+(b+d)^2-2(ac+bd)\ge 2(a+c)(b+d)-2(ac+bd) =2-2(ac+bd) $$ You now need to show $2[1-(ac+bd)]^2/(1+ac +bd)\ge 1/3$, which is equivalent to $ ac+bd\le 1/2$. Can you take it from here?

Answer 4

Just another way, we can use a lesser used but quite useful generalisation of the Radon inequality (see here for e.g.). Reproducing for ease of reference, for positive reals $a_i, b_i, r,s$ s.t. $r\geqslant s+1$ it says: $$\sum_{i=1}^n \frac{a_i^r}{b_i^s} \geqslant \frac{\left(\sum_i a_i \right)^r}{n^{r-s-1}\;\left( \sum_i b_i\right)^s}$$ It can now be easily noted that for our problem$$\sum \frac{a^3}{b+c+d}\geqslant \frac{(a+b+c+d)^3}{4\cdot3\cdot(a+b+c+d)}=\frac1{12}(a+b+c+d)^2$$ Now by AM-GM, $(a+b+c+d)^2\geqslant 4(a+c)(b+d)=4$ finishes this.