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.

Note that
$$
(|a|+|b|)^2=a^2+b^2+2|a||b|\leq a^2+b^2+a^2+b^2=2(a^2+b^2)
$$
and similarly for $(|c|+|d|)^2$.
