If $|a|+|b| > |c|+|d|$, is there must $a^2+b^2 > c^2+d^2$?

Question

If $|a|+|b| > |c|+|d|$, is there must $a^2+b^2 > c^2+d^2$?

Answer 1

No. Take $c=2,d=0.1$, $a=b=1.1$.

Answer 2

One shouldnt think so.

$|a| + |b| > |c| + |d| \implies$

$(|a| + |b|)^2 > (|c| + |d|)^2\implies$

$|a|^2 + 2|ab| + |b|^2 > |c|^2 + 2|cd| +|d|^2$

and that's a different equation altogether.

It's conceivable to have $|ab|$ significantly larger than $|cd|$ so that

$|a|^2 + |b|^2 = (|a|+|b|)^2 - 2|ab| < (|c|+|d|)^2 - 2|cd| =|c|^2 +|d|^2$.

Remember AM.GM.

$\frac {m+n}2 \ge \sqrt{mn}$ with equality holding if $m=n$ and with inequality accentuated if $m$ and $m$ are far apart!

So let $c= 1$ and $d=100$ and $c+d =101$ while $a=b=51$ so $a+b = 102$.

$a^2 + b^2 = 51^2 + 51^2 = 5,202$ while $c^2 + d^2 = 1+100^2 =10,001$