$(a^4+b^2) \leq (a^2+b^2)^2$ for all $a, b \in \mathbb{R}_+$? Let $a, b \in \mathbb{R}_+$.
Question. Is valid that
$$(a^4+b^2) \leq (a^2+b^2)^2?\tag{1}$$
I tried what follows:
$$(a^2+b^2)^2=a^4+2a^2b^2+b^4 \geq a^4+2a^2b^2.$$
For this, can I to conclude that $(1)$ holds?

If $b\ne 0$ then $b^2>0$ so $(b^2C\le b^2D\iff C\le D)$ for any $C,D.$
So with $C=1$ and $D=2a^2+b^2,$ if $b\ne 0$ then $$a^4+b^2\le (a^2+b^2)^2 \iff a^4+b^2\le a^4+2a^2b^2+b^4 \iff$$ $$\iff b^2\le 2a^2b^2+b^4\iff$$ $$\iff b^2(1)\le b^2(2a^2+b^2)\iff$$ $$\iff 1\le 2a^2+b^2.$$ The last inequality above cannot be true for every $a,b>0.$ E.g. if $a=b=1/1000000000000.$
