$a, b,$ and $c$ are not necessarily sides of a triangle, but they are positive numbers.

The question was:

Given that $a+b>c$ , does this imply that $\sqrt a +\sqrt b > \sqrt c$ , and that $a^2+b^2>c^2$ ?

The one with the square roots is easily proven by assuming it is true and squaring both sides. However, the one with the squares is giving be problems. I cannot arrive at a strong result. Of course sometimes $a^2+b^2$ will be equal to $c^2$ when $a, b$ are sides of a triangle, but I am not able to definitevly prove that $a^2+b^2$ is never $less$ than $c^2$ , so I am starting to believe that this statement is false. What is the way to solve this? Thanks.


2 Answers 2


Counterexample for the one with the squares:

$a = 6, b = 7, c = 12$


$a + b > c$


$a^2 + b^2 < c^2$

That counterexample is sufficient to disprove the conjecture.

Additionally, if $a + b > c$, then all 3 numbers can be thought of as the sides of a triangle. Then because of the law of cosines,

$a^2 + b^2 = c^2 - 2ab\cos(\gamma)$

The $\cos$ function can yield a positive, $0$, or negative result, so any of the following can be true:

  • $a^2 + b^2 < c^2$
  • $a^2 + b^2 = c^2$
  • $a^2 + b^2 > c^2$
  • $\begingroup$ Thanks, I suspected that it was possible to have $a^2+b^2$ less than, equal to, or greater than $c^2$ since I could never arrive at one definative result. But can you prove it algebraically? $\endgroup$
    – Ovi
    Jul 10, 2013 at 16:38
  • 2
    $\begingroup$ A counterexample is a proof that the statement is false. $\endgroup$
    – Sneftel
    Jul 10, 2013 at 16:41
  • $\begingroup$ The proof is good, but it assumes that $a, b, c$ $are$ the sides of a triangle. But if you had for example $a=20$ , $b=1$ , and $c=10$ , you couldn't have any sort of triangle. $\endgroup$
    – Ovi
    Jul 10, 2013 at 17:15

$a+b>c\Rightarrow a+b+2\sqrt{ab}>c\Rightarrow (\sqrt(a)+\sqrt{b})^2>(\sqrt{c})^2\Rightarrow \sqrt{a}+\sqrt{b}>\sqrt{c}$

Counterexample of the 2nd case, $0.5+0.5>0.9$ but $(0.5)^2+(0.5)^2=0.5<0.81$

  • $\begingroup$ Thanks, but I knew how to solve that. I was asking about showing if $a^2+b^2>c^2$ $\endgroup$
    – Ovi
    Jul 10, 2013 at 16:37

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