Find, in terms of $a$, where $a \gt 0$, the minimum value of the expression $\displaystyle \frac{a(x^2+y^2+c^2)+9xyz}{xy+yz+zx}$ for all non-negative real $x,\,y$ and $z$ such that $x+y+z=1$.

Find, in terms of $a$, where $a \gt 0$, the minimum value of the expression $\displaystyle \frac{a(x^2+y^2+c^2)+9xyz}{xy+yz+zx}$ for all non-negative real $x,\,y$ and $z$ such that $x+y+z=1$.

Since $9xyz≥4(xy+yz+zx)-1$ (by the Schur's inequality), we can transform the objective function as

$\displaystyle \frac{a(x^2+y^2+c^2)+9xyz}{xy+yz+zx}$

$\displaystyle =a\left(\frac{x^2}{xy+yz+zx}+\frac{y^2}{xy+yz+zx}+\frac{z^2}{xy+yz+zx}\right)+\frac{9xyz}{xy+yz+zx}$

$\displaystyle \ge a\left(\frac{x^2}{xy+yz+zx}+\frac{y^2}{xy+yz+zx}+\frac{z^2}{xy+yz+zx}\right)+\frac{4(xy+yz+zx)-1}{xy+yz+zx}$ (by the Schur's inequality)

$\displaystyle = a\left(\frac{x^2}{xy+yz+zx}+\frac{y^2}{xy+yz+zx}+\frac{z^2}{xy+yz+zx}\right)+4-\frac{1}{xy+yz+zx}$

$\displaystyle \ge a\left(\frac{(x+y+z)^2}{3(xy+yz+zx)}\right)+4-\frac{1}{xy+yz+zx}$ (by the extended Cauchy-Schwarz inequality)

$\displaystyle = a\left(\frac{1}{3(xy+yz+zx)}\right)+4-\frac{1}{xy+yz+zx}$

$\displaystyle = \frac{a-3}{3(xy+yz+zx)}+4$

$\displaystyle \ge \frac{a-3}{3\left(\frac{(x+y+z)^2}{3}\right)}+4$ since $\displaystyle xy+yz+zx\le \frac{(x+y+z)^2}{3}$

$\displaystyle = a-3+4$

$\displaystyle = a+1$

Therefore, the minimum of  $\displaystyle \frac{a(x^2+y^2+c^2)+9xyz}{xy+yz+zx}$ is $a+1$, this occurs when $x=y=z$.