Inequality

If $x,y$ are positive ,then by simply applying AM-GM 2 times , we get , $x^4+y^4 +8 \geq 2x^2y^2 +8 \geq 2\cdot 4xy=8xy$

Equality occurs at $x=y=\sqrt{ 2}$

Note that if $x,y$ are not positive, AM-GM inequality cannot be applied.

Rewrite $x^4+y^4+8\ge 8xy$ as $x^4+4+y^4+4\ge 8xy$

Now by $\color{#3D99F6}{\text{AM-GM inequality}}$ on $x^4+4$ we get

$\begin{aligned} \dfrac{x^4+4}{2}\ge \sqrt{x^4\cdot 4}\\& \implies x^4+4\ge 2(2x^2)\\& \implies x^4+4\ge 4x^2 ~-~ \text{(i)}\end{aligned}$

Again by $\color{#3D99F6}{\text{AM-GM inequality}}$ on $y^4+4$ we have

$\begin{aligned} \dfrac{y^4+4}{2}\ge \sqrt{y^4\cdot 4} \\& \implies y^4+4\ge 2(2y^2) \\& \implies y^4+4\ge 4y^2 ~-~ \text{(ii)} \end{aligned}$

Now by $\text{(i)}+\text{(ii)}$

$x^4+y^4+8\ge 4x^2+4y^2 ~-~ \text{(iii)}$

now by using $\color{#3D99F6}{\text{AM-GM inequality}}$ on the R.H.S of $\text{(iii)}$ we have

$\begin{aligned} \dfrac{4x^2+4y^2}{2}\ge \sqrt{4x^2\cdot 4y^2} \\& \implies 4x^2+4y^2\ge 2\cdot 4xy\\& \implies 4x^2+4y^2\ge 8xy\end{aligned}$

Since we have $x^4+y^4+8\ge 4x^2+4y^2$ and $4x^2+4y^2\ge 8xy$

Hence, $x^4+y^4+8\ge 8xy$

$\therefore$ the inequality is TRUE

$\frac{x^{4}+y^{4}+4+4}{4} \geq (4 \cdot 4 \cdot x^4 \cdot y^4)^{1/4}$