$(a^4+b^2) \leq (a^2+b^2)^2$ 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?
 A: $$\bigg( \left(\frac{1}{2} \right)^2+ \left(\frac{1}{2} \right)^2 \bigg)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}$$
but
$$\bigg( \left(\frac{1}{2}\right)^4+ \left(\frac{1}{2}\right)\bigg)^2= \left(\frac{1}{2}\right)^4+\frac{1}{4} > \frac{1}{4}= \bigg( \left(\frac{1}{2} \right)^2+ \left(\frac{1}{2}\right)^2 \bigg)^2$$
A: HINT
You can simply expand the RHS and subtract the LHS:
\begin{align*}
(a^{2} + b^{2})^{2} = a^{4} + 2a^{2}b^{2} + b^{4} \Rightarrow (a^{2} + b^{2})^{2} - a^{4} - b^{2} = 2a^{2}b^{2} + b^{4} - b^{2} = b^{2}(2a^{2} + b^{2} - 1)
\end{align*}
When the last expression is nonnegative?

 Answer: outside the ellipse $2a^{2} + b^{2} = 1$

A: $a=0, b=\frac 1 2 $ is a counter-example.
If $0$ is not allowed just take $a>0$ but small enough. Eg: $a=\frac 1 {10}$ will do.
A: 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 $a=b=1/1000000000000.$
