(hint: show that a+b>c)
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This one is easiest to approach using proof by contradiction: suppose a + b <= c and show this forces one of the hypotheses to fail.
Now what to do with that? How about squaring both sides. You get
a^2 + 2ab + b^2 <= c^2. Using the second hypothesis and subtracting equal quantities from both sides gives 2ab <= 0, but that is impossible if a, b > 0. Therefore our supposition is false and we have a + b > c.
Now that doesn't finish the problem, you still need to see that a + c > b and b + c > a for there to be an
actual triangle, but those should be easier.
Now what to do with that? How about squaring both sides. You get
a^2 + 2ab + b^2 <= c^2. Using the second hypothesis and subtracting equal quantities from both sides gives 2ab <= 0, but that is impossible if a, b > 0. Therefore our supposition is false and we have a + b > c.
Now that doesn't finish the problem, you still need to see that a + c > b and b + c > a for there to be an
actual triangle, but those should be easier.