# Twin Inequalities in Four Variables: Twin 2

### Solution

After expanding and simplifying, the required inequality becomes

$\displaystyle a^{\frac{5}{3}}d^{\frac{5}{3}}b^{\frac{1}{3}}c^{\frac{1}{3}}+b^{\frac{5}{3}}c^{\frac{5}{3}}a^{\frac{1}{3}}d^{\frac{1}{3}}\le a^2d^2+b^2c^2.$

Let $x=a^{\frac{1}{3}},\,$ $y=b^{\frac{1}{3}},\,$ $u=c^{\frac{1}{3}},\,$ $v=d^{\frac{1}{3}}.\,$ The new variables are all positive. We need to prove that

$\displaystyle F=x^5v^5yu+y^5u^5xv-x^6v^6-y^6u^6\le 0.$

We have

\begin{align} F&=x^5v^5yu+y^5u^5xv-x^6v^6-y^6u^6\\ &=(yu-xv)\left((xv)^5-(yu)^5\right)\\ &=-(yu-xv)^2\left(\sum_{k=0}^4(xv)^k(yu)^{4-k}\right)\le 0, \end{align}

with equality when $yu=xv,\,$ i.e., $a^{\frac{1}{3}}d^{\frac{1}{3}}=b^{\frac{1}{3}}c^{\frac{1}{3}},\,$ or $ad=bc.$

### Acknowledgment

Dan Sitaru has kindly posted this problem from the Romanian Mathematics Magazine at the CutTheKnotMath facebook page, along with three solutions. The above solution is by Soumava Chakraborty; Ravi Prakash and Seyran Ibrahimov have independently submitted two solutions along the same lines.