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Prove that at least one of the following:

$$4a-b^2$$

$$4b-c^2$$

$$4c-a^2$$

is less than or equal to $4$.

Any help in this would be really helpful! I have been stuck on this for a while. So far my attempt has been assuming that they are all greater than $4$ and seeing if I can get a contradiction, but I have not made much progress here...

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You are going on the right track. Assume that all three expressions are greater than $4$.

Then we will have $4a-b^2+4b-c^2+4c-a^2> 12$

$\Leftrightarrow a^2-4a+b^2-4b+c^2-4c+12< 0$

$\Leftrightarrow a^2-4a+4+b^2-4b+4+c^2-4c+4< 0$

$\Leftrightarrow (a-2)^2+(b-2)^2+(c-2)^2< 0$

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