Proving $(|a|+|b|)^2(|c|+|d|)^2 \leq 4(a^2+b^2)(c^2+d^2)$. Given $a,b,c,d \in \Bbb R$, I want to prove that: $$(|a|+|b|)^2(|c|+|d|)^2 \leq 4(a^2+b^2)(c^2+d^2).$$
But the bound I'm getting is just too weak. I know that: $$|a| = \sqrt{a^2} \leq \sqrt{a^2+b^2} \implies |a|+|b| \leq 2\sqrt{a^2+b^2}\implies (|a|+|b|)^2\leq 4(a^2+b^2).$$ This gives me:  $(|a|+|b|)^2(|c|+|d|)^2 \leq 16(a^2+b^2)(c^2+d^2)$. Not good. I tried reverse engineering it: $$\begin{align} (|a|+|b|)^2 &\leq 2(a^2+b^2) \\ |a|+|b| &\leq \sqrt{2}\sqrt{a^2+b^2} \\ |a|+|b| &\leq \sqrt{2a^2+2b^2} \\ |a|+|b|&\leq  \sqrt{(a\sqrt{2})^2+(b\sqrt{2})^2},\end{align}$$ which suggests that I may try to prove that: $$|a| \leq \frac{\sqrt{2}}{2}\sqrt{a^2+b^2}.$$
I'm failing to see the step needed here. I just need a little push. Thanks.

Here's another way of proving it. From the Cauchy-Schwarz Inequality we have:
$$(1 + 1)(a^2 + b^2) \ge \left(\sqrt{1\cdot a^2} + \sqrt{1\cdot b^2}\right)^2 = \left(\mid a \mid + \mid b \mid\right)^2$$
$$(1 + 1)(c^2 + d^2) \ge \left(\sqrt{1\cdot c^2} + \sqrt{1\cdot d^2}\right)^2 = \left(\mid c \mid + \mid d \mid\right)^2$$
Now multiply them to get:
$$4(a^2 + b^2)(c^2 + d^2) \ge (\mid a \mid + \mid b \mid)^2(\mid c \mid + \mid d \mid)^2$$
Hence the proof. We have equality when:
$$\frac {a^2}1 = \frac {b^2}1 \quad \text{ and } \quad \frac {c^2}1 = \frac {d^2}1$$
Or in other words when $a=b$ and $c=d$
