Twin Inequalities in Four Variables: Twin 1

Solution 1

Recollect the Cauchy-Schwarz inequality:

For real numbers $x_1,x_2,\ldots,x_n\,$ and $y_1,y_2,\ldots,y_n\,$ and integer $n\ge 1,$

$\displaystyle \left(\sum_{k=1}^nx_ky_k\right)^2\le\left(\sum_{k=1}^nx_k^2\right)\left(\sum_{k=1}^ny_k^2\right),$

with equality when the vectors $\mathbf{x}=(x_1,\ldots,x_n)\,$ and $\mathbf{y}=(y_1,\ldots,y_n)\,$ are proportional.

Take $x_1=\sqrt{b}\sqrt[10]{ab^4},\,$ $y_1=\sqrt{a}\sqrt[10]{a^4b},\,$ $x_2=\sqrt{d}\sqrt[10]{cd^4},\,$ $y_2=\sqrt{c}\sqrt[10]{c^4d}.\,$ Then the above becomes

$(ac+bd)^2\le\left(b\sqrt[5]{ab^4}+d\sqrt[5]{cd^4}\right)\left(a\sqrt[5]{a^4b}+c\sqrt[5]{c^4d}\right),$

with equality when $\displaystyle \frac{\sqrt{b}\sqrt[10]{ab^4}}{\sqrt{a}\sqrt[10]{a^4b}}=\frac{\sqrt{d}\sqrt[10]{cd^4}}{\sqrt{c}\sqrt[10]{c^4d}},\,$ i.e., for $bc=ad.$

Solution 2

Let $a=x^5,\,$ $b=y^5,\,$ $c=z^5,\,$ $d=t^5.\,$ Then the required inequality becomes

$\displaystyle (x^5y^5+z^5t^5)^2\le (y^9x+t^9z)(x^9y+z^9t)$

which is, when expanded and simplified, reduces to

$\displaystyle xty^9z^9+zyx^9t^9\ge 2x^5y^5z^5t^5$

which is true by the AM-GM inequality.

Acknowledgment

Dan Sitaru has kindly posted this problem from the Romanian Mathematics Magazine at the CutTheKnotMath facebook page, along with three solutions. Solution 1 is by Kevin Soto Palacios; Chris Kyriazis independently submitted a simialr solution; Solution 2 is by Seyran Ibrahimov.