Proving whether or not there exist $a,b$ such that $a+b>100$ and $a^2+b^2<1/1000$

Question

sorry for my English unfortunately I don't know English so well

Prove if the statement is true or false: For two real positive numbers a,b: a+b>100 and a^2+b^2<1/1000

I know the answer is false by instinct but I don't know how to prove it..

Answer 1

Your intuition should tell you that if $a > 0; b>0$ then $a^2 > 0$ and $b^2 > 0$ so $a^2 < a^2 + b^2 < \frac 1{1000} < 1$ so $a < 1$ and $b^2 < a^2 + b^2 < \frac 1{1000} < 1$ so $b < 1$ so $a + b < 1 + 1 < 2$ and $a + b > 100$ is not possible.

There are no solutions.

More formally: If $a + b > 100$ then $avg(a,b) = \frac {a + b}2 > 50$. Let $avg(a,b) = c$ and let $k = c - a = \frac {a+b}2 - a = \frac {b -a }2$ so $a = c - k$ and $b - k = b - \frac {b-a}2 = \frac {a+b}2=c$ so $b = c + k$.

So $a^2 + b^2 = (c-k)^2 + (c + k)^2 = c^2 - 2ck + k^2 + c^2 +2ck + k^2 = 2c^2 +k^2 > 2c^2 > 2*50^2 = 5000> \frac 1{1000}$

So $a^2 + b^2 < \frac 1{1000}$ is most certainly impossible.

Answer 2

Since a + b > 100, either a or b has to be greater than 1.

Therefore, either a^2 or b^2 has to be greater than 1.

Hence, the sum a^2 + b^2 cannot be less than 1/1000