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For any $a$, $b$, and $c$ prove $$3a^2+3b^2-2b+2a+1>0$$ I tried the following $$(a+1)^2+(b-1)^2+2(a^2+b^2)-1>0\\ (a+1)^2-1+(b-1)^2+2(a^2+b^2)>0\\ (a+1-1)(a+1+1)+(b-1)^2+2(a^2+b^2)>0\\ (a^2+2a)+(b-1)^2+2(a^2+b^2)>0\\ $$ $a^2$ and $(b-1)^2$ and $2(a^2+b^2)$ are always positive and greater than zero, when $a>0$ our inequality is proved. However when a is negative I can not say $2a$ is always less than or equal to $a^2$. I think this problem should be solved by algebraic identities alone and not by discussing the different signs of $a$? I appreciate your help

user2345215
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max
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  • max, I deleted your "thank you" post. You can thank for the help you received by accepting and upvoting. See this bit in our help section. You are a new user here, and you probably didn't know about this. You may benefit from reading other parts of the help section as well. Alternatively you can just roam here, and see how we do things at Math.SE. Welcome to Math.SE from me, too. Hope you enjoy the site. – Jyrki Lahtonen Dec 25 '14 at 19:11

3 Answers3

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$$3a^2+3b^2−2b+2a+1 = (3 a^2 + 2a +\frac{1}{3}) + (3 b^2 - 2b + \frac{1}{3}) + \frac{1}{3} = $$ $$ = (\sqrt{3}a + \frac{1}{\sqrt{3}})^2 + (\sqrt{3}b - \frac{1}{\sqrt{3}})^2 + \frac{1}{3} > 0 $$

Jakubussimus
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Hint: $$(a-b+1)^2=a^2+b^2+1-2ab-2b+2a$$


Solution: $$3a^2+3b^2-2b+2a+1=2a^2+2b^2+(a-b+1)^2+2ab=a^2+b^2+(a+b)^2+(a-b+1)^2>0$$

user2345215
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$f(a) = 3a^2 +2a + (3b^2-2b+1) \to \triangle' = 1^2 - 3(3b^2-2b+1) = -9b^2 +6b - 2 < 0, \forall b \text{ since } \triangle' = 3^2 -(-9)(-2) = 9 - 18 = -9 < 0 \to f(a) > 0, \forall a$.

DeepSea
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