# An Inequality with Two Triples of Variables II

### Solution

The required inequality can be rewritten as

$\displaystyle 3\sqrt{\left(\sum_{cycl}a^2\right)\left(\sum_{cycl}x^2\right)}\ge x(2b+2c-a)+y(2c+2a-b)+z(2a+2b-c).$

Set $2b+2c-a)=3u,$ $2c+2a-b=3v,$ $2a+2b-c=3w.$ Then $a^2+b^2+c^2=u^2+v^2+w^2,$ and the inequality becomes

$\displaystyle \sqrt{\left(\sum_{cycl}a^2\right)\left(\sum_{cycl}x^2\right)}\ge ux+vy+wz,$

which is just the Cauchy-Schwarz inequality.

### Solution 2

\displaystyle \begin{align} &\frac{2}{3}\left(\sum_{cycl}a\right) \left(\sum_{cycl}x\right)-(ax+by+cz) \\ &=\frac{1}{3}\left[a(2y+2z-x)+b(2z+2x-y)+c(2x+2y-z)\right] \\ &\leq\frac{1}{3}\sqrt{\left(\sum_{cycl}a^2\right)\left[\sum_{cycl}(2y+2z-x)^2\right]}~\text{(CS)} \\ &=\frac{1}{3}\sqrt{\left(\sum_{cycl}a^2\right) \left(9\sum_{cycl}x^2\right)}=\sqrt{\left(\sum_{cycl}a^2\right) \left(\sum_{cycl}x^2\right)}. \end{align}

### Acknowledgment

I am grateful to Leo Giugiuc for mailing me this problem, along with a solution of his. The problem was posted at Lorian Saceanu's facebook group Easy beautiful math with a reference to Kvant (1989) and credits to Vasile Cirtoaje. Solution 2 is by Amit Itagi.