<s>[INST] Suppose that a and b are real numbers with $0< a <1 $ and $0< b <1$. Prove by contradiction or contrapositive: If $a^2+b^2= 1$, then $a+b >1$. So far I get the below:
To prove the above statement with contrapositive, we need to show that if $a+b \leq 1$ then $a^2 + b^2 \ne 1$. If $a+b \leq 1$, then $a \leq 1 - b$. If we square both sides of the inequality we get $a^2 \leq (1 - b)^2$, which is $a^2 \leq 1 - 2b + b^2$, then $a^2 - b^2 \leq 1 - 2b$.
I get to this step but I get $a^2 - b^2$ instead of $a^2 + b^2$, I don't know if I am on the right track or if it is possible to prove with contrapositive?
 [/INST] If $$a+b \le 1$$ upon squaring both sides we get, $$a^2+b^2+2ab\le 1$$
That implies $$1+2ab\le1$$ or $$2ab\le 0$$
Which is impossible due to $0<a<1$ and $0<b<1$
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