$a^2+b^2\le c^2+d^2$ where $a<c$ and $b>d$

Question

I want to show that $a^2 + b^2 \le c^2 + d^2$ where $a < c$ and $b>d$ and $a, b, c, d$ are strictly between 0 and 1 and $c \ne d$ and $a \ne b$. How can I do this? Is it possible to prove this? I will be grateful for any help.

Answer 1

First you should observe that $0 < x < 1$ if and only if $x$ is positive and $0 < x^2 < 1$. (Similarly $x < y$ is equivalent to $x^2 < y^2$, for positive numbers $x$ and $y$.) So there is no reason to use the squared terms. Then restate the question:

$\quad$ Claim (false): If $0 < a < c < 1$ and $0 < d < b < 1$, then $a + b\leq c + d$.

Can you see some counterexamples now? (There are many.) More dramatically, since the roles of $a,c$ and $b, d$ are symmetric, if the claim were true then we would have to conclude that $a + b = c + d$ for any such $a,b,c,d$!