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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?

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  • $\begingroup$ Add $b^2$ to both sides of $a^2 \leq 1 - 2b + b^2$. Then you have "$a^2 + b^2 \leq 1 - (\text{stuff involving only $b$})$. If you can show the "(stuff)" is positive, you're done because one minus a positive amount is strictly less than one. I might factor the (stuff). $\endgroup$ Oct 30, 2021 at 1:03
  • $\begingroup$ You can write $a=\cos \varphi $ and $b=\sin \varphi $ with $0<\varphi <\pi/2.$ This gives you a differentiable function $f(\varphi)=\sin(\varphi )+\cos(\varphi ) $ and you can prove that it is greater than $1$ on the domain $0<\varphi <\pi/2.$ $\endgroup$ Oct 30, 2021 at 1:06

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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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If you start with $a+b\leq1$ then $b\leq1-a$. We also have by hypothesis that $a^2+b^2 = 1$, which means that: $$a^2 + b^2 = 1$$ $$a^2 + (1-a)^2 \geq 1$$

That can be written: $$2a^2-2a\geq0$$ So: $$2a(a-1)\geq0$$ To be true, this means that $a = 0$ or $a\geq1$.

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Using geometry (Pythagoras) : a rectangular triangle with hypotenuse 1 and other sides a and b. For sure: a + b > 1 (as is so for any 'non flat' triangle).

Agreed: not really proof by contradiction or contrapositive (except in that a rectangular triangle is never 'flat')

discrete mathematics is perhaps not optimal adequate tag.

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