Find Min of $\frac{a(x^2+y^2+z^2)+9xyz}{xy+yz+zx}$ for $x+y+z=1$

[anemone]

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

 

[kaliprasad]

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

Correction: it should be z instead of c

Spoiler

applying cyclic symmetry the value is at $x=y=z= \frac{1}{3}$ and the value is a + 1 . the value at $(1,0,0)$ is infinite

 

[anemone]

Your answer is correct, thanks for participating, kaliprasad!

Spoiler

I hope someone else could solve it without using the property of cyclic symmetry. :)

 

[Albert1]

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

Using $AP\geq GP$

Spoiler

$\dfrac {a(x^2+y^2+z^2)+9xyz}{xy+yz+zx}
\geq\dfrac {a(x^2+y^2+z^2)}{x^2+y^2+z^2}+\dfrac {9xyz}{xy+yz+zx}$
$\geq a+27xyz=a+1$
(equality holds at $x=y=z=\dfrac {1}{3}$

 

[anemone]

Using $AP\geq GP$

Spoiler

$\dfrac {a(x^2+y^2+z^2)+9xyz}{xy+yz+zx}
\geq\dfrac {a(x^2+y^2+z^2)}{x^2+y^2+z^2}+\dfrac {9xyz}{xy+yz+zx}$
$\geq a+27xyz=a+1$
(equality holds at $x=y=z=\dfrac {1}{3}$

Thanks Albert for participating! Your approach is correct and neater than mine!(Cool)

My solution:

Spoiler

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

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

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

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

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

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

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

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

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

$$= a-3+4$$

$$= a+1$$

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