# 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?

• 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). Oct 30, 2021 at 1:03
• 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 $0<\varphi <\pi/2.$ Oct 30, 2021 at 1:06

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 $$0 and $$0
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 $$a\geq1$$.