Find, in terms of $a$, where $a \gt 0$, the minimum value of the expression [MATH]\frac{a(x^2+y^2+c^2)+9xyz}{xy+yz+zx}[/MATH] 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

[MATH]\frac{a(x^2+y^2+c^2)+9xyz}{xy+yz+zx}[/MATH]

[MATH]=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}[/MATH]

[MATH]\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}[/MATH] (by the Schur's inequality)

[MATH]= 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}[/MATH]

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

[MATH]= a\left(\frac{1}{3(xy+yz+zx)}\right)+4-\frac{1}{xy+yz+zx}[/MATH]

[MATH]= \frac{a-3}{3(xy+yz+zx)}+4[/MATH]

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

[MATH]= a-3+4[/MATH]

[MATH]= a+1[/MATH]

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

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