Non-Symmetric Inequality

$\begin{aligned} \frac{2x^4+4y^4+9}{2x^2+y^2} & = \frac{2x^4 + 9\color{#3D99F6}{x^2 y^2} + 4y^4}{2x^2+y^2} \quad \quad \color{#3D99F6}{[xy=1]} \\ & = \frac{(2x^2 + y^2)(x^2 + 4y^2)}{2x^2+y^2} \\ & = \color{#3D99F6} {x^2 + 4y^2 \ge 2\sqrt{4x^2y^2}} = 4xy = \boxed{4} \quad \quad \color{#3D99F6}{[AM \ge GM]} \end{aligned}$

First is to factor left hand side

$2x^{4}+4y^{4}+9=(2x^{2}+y^{2})(4y^{2}+x^{2})$

So,

$\dfrac{2x^{4}+4y^{4}+9}{2x^{2}+y^{2}}=4y^{2}+x^{2}$ Since x and y aren't equal to 0.

use $x=\dfrac{1}{y}$

then we have

$4y^{2}+\dfrac{1}{y^{2}}$ and we have to find the minimum of this funtion.

Since the derivative must equal to 0.

$8y-\dfrac{2}{y^{3}}=0$ then $y^{4}=\dfrac{1}{4},y^{2}=\dfrac{1}{2}$

Hence

$4y^{2}+\dfrac{1}{y^{2}}=2+2=\boxed{4}$

A different solution:

The minimum of the expression is the greatest k such that

$\frac{2 x^4+4 y^4+9}{2 x^2+y^2}\geq k$

for all nonzero x,y such that x y = 1. Re-arranging, we must have that

$\frac{1}{2} \left(x^2-\frac{k}{2}\right)^2+\left(y^2-\frac{ k}{8}\right)^2\geq \frac{1}{4} \left(\frac{9 k^2}{16}-9\right)$

Note that if $-4<k<4$ then the above inequality is true for all x,y. Hence, k>=4.

Suppose k>4. Then we cannot find x,y with xy=1 so that the expression equals 4. However, if (x,y)=(sqrt(2),1/sqrt(2)), then the expressin equals 4. Hence, we have a contradiction. So k=4.

Using Cauchy-Schwartz Inequality and then AM-GM inequality, $\left(2+\frac{1}{4}+\frac{9}{4}\right)\left(\frac{2x^4+4y^4+9}{2x^2+y^2}\right) \ge \frac{\left(2x^2+y^2+\frac{9}{2}\right)^2}{2x^2+y^2}=9+(2x^2+y^2)+\frac{81}{4(2x^2+y^2)} \ge 9+2\left(\frac{9}{2}\right)=18$

Hence, $\frac{2x^4+4y^4+9}{2x^2+y^2} \ge 18\times \frac{2}{9}=4$

The equality holds if and only if $\frac{2x^4}{2}=\frac{4y^4}{\frac{1}{4}}=\frac{9}{\frac{9}{4}}$ and $2x^2+y^2=\frac{9}{2}$ . It is possible if $(x,y)=\left(\sqrt{2},\frac{1}{\sqrt{2}}\right)$ .

Let $\frac{2x^4 +4y^4 +9}{2x^2 +y^2} \geq\ C$ using xy = 1 (to make the inequality homogeneous) and rearranging $\Rightarrow\ 2x^4 +4y^4 +9x^2 y^2 \geq\ C (2x^2 +y^2)$ Factorising the LHS: $\ (2x^2 +y^2)(x^2 +4y^2) \geq\ C(2x^2 +y^2)$ $\ (2x^2 +y^2)(4y^2 +x^2 -C) \geq\ 0$ Using xy = 1 again: $\ (2x^2 +y^2)(4y^2 +x^2 -Cxy) \geq\ 0$ Now we need the 2nd bracket to be a square... $\ 4y^2 +x^2 -Cxy \equiv\ (a-b)^2 \Rightarrow\ ( 4y^2 +x^2 -Cxy) = (2y-x)^2$ $\therefore\ Min \ Value = C = 4$ This is achieved when $\ x = 2y \Rightarrow\ xy = 2y^2 = 1 \Rightarrow\ y= \pm \frac{\sqrt{2}}{2} \ and \ x = \pm \sqrt{2}$