Extremas of a Constrained Inequality!

Let $P(x,y,z)=\dfrac{x^4+y^4+z^4}{(x+y+z)^4}$ .

Note that $P(x, y, z)$ is homogeneous, that is $P(tx, ty, tz)=P(x,y,z),\forall t>0$ .

Also note that if $(x, y, z)$ satisfy conditions of the problem, then so does $(tx, ty, tz),\;t>0$ .

Therefore, without loss of generality, we can assume that $x+y+z=4$ and hence $xyz=2$ .

So the problem can restated as follows: Find the maximum and minimum values of $P=\dfrac{1}{256}(x^4+y^4+z^4)$ if $x, y, z>0$ satisfying $x+y+z=4$ and $xyz=2$ .

Put $A=x^4+y^4+z^4, B=xy+yz+zx$ , we have:

$A=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+z^2x^2)$

$\quad=((x+y+z)^2-2(xy+yz+zx))^2-2[(xy+yz+zx)^2-2xyz(x+y+z)]$

$\quad=2(B^2-32B+144)$

By the condition $x+y+z=4$ and $xyz=2$ , the inequality $(y+z)^2\ge4yz$ is equivalent to $(4-x)^2\ge\dfrac{8}{x} \Leftrightarrow 3-\sqrt5\le x\le 4\quad$ (as $0\le x\le 4)$ .

By symmetry, we also have $3-\sqrt5\le y,z\le 2$ .

Then we have:

$(x-2)(y-2)(z-2)\le 0 \Leftrightarrow 5\le B$ .

$(x-(3-\sqrt5))(y-(3-\sqrt5))(z-(3-\sqrt5))\ge 0 \Leftrightarrow B\le \dfrac{5\sqrt5-1}{2}$ .

Since $A=f(B)=2(B^2-32B+144)$ , as a quadratic function of $B$ , is decreasing on $(0, 16)\supset \left[5; \dfrac{5\sqrt5-1}{2}\right]$ , we obtain

$\min A=f\left(\dfrac{5\sqrt5-1}{2}\right)=383-165\sqrt5; \max A=f(5)=18$ .

Thus,

$\min P=\dfrac{383-165\sqrt5}{256}$ , occurs at say $x=3-\sqrt5, y=z=\dfrac{1+\sqrt5}{2}$

$\max P=\dfrac{9}{128}$ , occurs at say $x=2; y=z=1$

So, the minimum value of $A+B+C+D+E+F$ is $383+165+5+256+9+128=\boxed{946}$ .

$\frac{x^{4} + y^{4} + z^{4}}{(x + y + z)^{4}}$ is homogeneous.

minimum or maximum values always occur when $x = y = z$ .

but $x = y = z$ does not satisfy $(x + y + z)^{3} = 32xyz$ .

So minimum or maximum values occur when $x = y$ or $y = z$ or $z = x$ .

Let $y = z = 1$ ,

$(x + 2)^{3} = 32x$ .

$x = 2, x = 2\sqrt{5} - 4, x = -2\sqrt{5} - 4 < 0$

$x = 2, y = z = 1$

$\frac{x^{4} + y^{4} + z^{4}}{(x + y + z)^{4}} = \frac{9}{128}$

$x = 2\sqrt{5} - 4, y = z = 1$

$\frac{x^{4} + y^{4} + z^{4}}{(x + y + z)^{4}} = \frac{383 - 165\sqrt{5}}{256}$

$\frac{383 - 165\sqrt{5}}{256} < \frac{9}{128}$

maximum value = $\frac{9}{128}$

minimum value = $\frac{383 - 165\sqrt{5}}{256}$

The minimum value of $A + B + C + D + E + F = 383 + 165 + 5 + 256 + 9 + 128 = \boxed{946}$