Fairly Interesting Inequality

$\begin{aligned} x^2+4xy+4y^2+2z^2 ~~&=& ~x^2+2xy+2xy+4y^2+z^2+z^2 \\ \\ & \stackrel{\text{AM GM}}{\ge} & ~6\sqrt[6]{x^2\cdot 2xy\cdot 2xy\cdot 4y^2\cdot z^2\cdot z^2} \\ \\ &=& ~6\sqrt[6]{2^4 (xyz)^4} \\ \\ &\stackrel{xyz~=~32}{=}& ~6\sqrt[6]{2^4 \cdot 2^{20}} \\ \\ &=& ~\boxed{96} \end{aligned}$ Equality occurs at $(x,y,z)=(4,2,4)$ , so $96$ indeed is the minimum.

$(x-2y)^2\ge0\implies x^2+4xy+4y^2+2z^2\ge8xy+2z^2$ $(z-4)^2\ge0\implies8xy+2z^2\ge8xy+16z-32$ $8(\sqrt{xy}-\sqrt{2z})^2\ge0\implies8xy+16z-32\ge96$ $x^2+4xy+4y^2+2z^2\ge96$ $4^2+4\cdot4\cdot2+4\cdot2^2+2\cdot4^2=\boxed{96}$

Using AM-GM inequality,

$x^2+2xy+2xy+4y^2+z^2+z^2 \geq 6 \times \sqrt[6]{2*2*4*(xyz)^4}$

$\Rightarrow x^2+4xy+4y^2+2z^2 \geq 6 \times \sqrt[6]{2*2*2^2*(2^5)^{4}}$

$\Rightarrow x^2+4xy+4y^2+2z^2 \geq \boxed{96}$

And the equality occurs at $(x,y,z) \equiv (4,2,4)$ .

Here's a technique for splitting the terms while applying AM-GM inequality. (I find this quite akin to balancing a chemical equation.)

Let each of the terms be split into a, b, c and d parts respectively.

$x^2 + 4xy + 4y^2 + 2z^2 = \dfrac{a}{a}x^2 + \dfrac{b}{b}4xy + \dfrac{c}{c}4y^2 + \dfrac{d}{d}2z^2$

For using AM-GM inequality, the powers of x, y and z should be the same in the product of the terms. So, $\underbrace{2a + b}_{\text{Power of } x} = \underbrace{2c + b}_{\text{Power of } y} = \underbrace{2d}_{\text{Power of } z}$ One possible solution is $a = c = 1, b = d= 2$

Thus, the expression becomes: $x^2 + \dfrac{4xy}{2} + \dfrac{4xy}{2} + 4y^2 + \dfrac{2z^2}{2} + \dfrac{2z^2}{2}$

Using AM-GM Inequality, $AM ≥ GM$ $\dfrac{x^2 + \dfrac{4xy}{2} + \dfrac{4xy}{2} + 4y^2 + \dfrac{2z^2}{2} + \dfrac{2z^2}{2}}{6} ≥ \sqrt[6]{16 x^4y^4z^4}$ $x^2 + 4xy + 4y^2 + 2z^2 ≥ 6 \sqrt[6]{2^4 32^4}$ $x^2 + 4xy + 4y^2 + 2z^2 ≥ 96$ So min value $= \boxed{96}$ (when $x = z= 4, y = 2$ )

There you go....Great question....Demanding quite some creativity....Post a few more harder problems..

$f(x,y)=x^2+2xy+4y^2+2z^2 = (x+2y)^2+2z^2=(x+2y)^2+\frac {{32}^2} {x^2y^2};$

we have imposed $xyz=32;\;z=\frac {32}{xy};$

$grad(f) = 0,\;$ which leads to:

$\begin{cases}x+2y=\frac {2^{11}}{x^3y^2}\\x+2y=\frac {2^{10}}{x^2y^3}\end{cases}\;(1)$

equating 2nd members of the two equations of $(1)\;$ , because 1st members are equal:

$2x^2y^3=x^3y^2;\;x=2y;$ substituting in 1st equation of $(1)\;:$

$x^6=2^{12};\;x^*=\pm 4;\;y*=\pm 2$

$f(x,y)\;$ is sum of all positive values, so has no maximum, so $(x^*,y^*)\;$ are points of minima, and the minimum is:

$f(x^*,y^*)=\boxed {96}$

We can also break the expression as

${ x }^{ 2 }+2xy+2xy+4{ y }^{ 2 }+{ z }^{ 2 }+{ z }^{ 2 }$

And then apply AM-GM inequality to it we get

${ x }^{ 2 }+4xy+4{ y }^{ 2 }+2{ z }^{ 2 }\ge 96$

Since $x^2 + 4xy + 4y^2 + 2z^2 = x^2 + 2xy + 2xy + 4y^2 + z^2 + z^2$ and by AM-GM inequality :

$x^2 + 2xy + 2xy + 4y^2 + z^2 + z^2 \ge 6 \sqrt[6]{16x^4y^4z^4} = \boxed{96}$

Equality occurs when $x=z=4$ and $y=2$ .

We have (x + 2y)^2 ≥ 8xy with equality iff x = 2y. Hence (x + 2y)^2 + 2z^2 ≥ 256/z + 2z^2 with equality iff x = 2y. We have z^2 + 128/z - 48 = (z - 4)^2(z + 8)/z ≥ 0 with equality iff z = 4 (for positive z). Hence 256/z + 2z^2 ≥ 96 with equality iff z = 4. So the expression given has minimum value 96, achieved only at x = 4, y = 2, z = 4.