I don't know how to prove that the following inequality holds (under condition $a+b+c=0$): $\frac{(2a+1)^2}{2a^2+1}+\frac{(2b+1)^2}{2b^2+1}+\frac{(2c+1)^2}{2c^2+1}\geqq 3$
Inequality under condition $a+b+c=0$
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0I found it as an exercise in one local book, but there isn't any solution for it; i've tried playing with it by adding multiples of a$+$b+c to both sides of the inequality and then modifying the fractions, but i was unable to advance in any way. – 2012-09-21
2 Answers
This one made me struggle so much that I was close to go crazy. Therefore let me post my solution to have a relief from this burden..
First we have $2a^2=\frac43a^2+\frac23a^2=\frac43a^2+\frac23(b+c)^2\leq \frac43(a^2+b^2+c^2);$ where the last inequality follows from the arithmetic-quadratic mean.
Analogously $\begin{split}2b^2&\leq \frac43 (a^2+b^2+c^2),\\ 2c^2&\leq \frac43(a^2+b^2+c^2).\end{split}$ It follows that $\sum_\text{cyc}\frac{(2a+1)^2}{2a^2+1}\geq3\sum_\text{cyc}\frac{(2a+1)^2}{4(a^2+b^2+c^2)+3}=3\left(1+\frac{4(a+b+c)}{4(a^2+b^2+c^2)+3}\right)=3.$
The following proof, as it is, only works for $a,b,c \not\in (-2,0)$.
Let $|a| \geq |b| \geq |c|$ $ \frac{(2x+1)^2}{2x^2+1} = 1 + \frac{2x^2+4x}{2x^2+1} $
Then it follows that
$ \text{left-hand side} = 3 + \frac{2a^2+4a}{2a^2+1} + \frac{2b^2+4b}{2b^2+1} + \frac{2c^2+4c}{2c^2+1} \geq 3 + \frac{2(a^2+b^2+c^2)+4(a+b+c)}{2a^2+1} = $
$ = 3 + \frac{2(a^2+b^2+c^2)}{2a^2+1} \geq 3 $
Now in the second inequality I assumed for $x\in\{a,b,c\}$ that $0 \leq 2x^2+4x = 2(x^2+2x) = 2x(x+2)$. As the parabola defined by $2x(x+2)$ only is negative between -2 and 0 exclusive we can do so savely if none of $a,b,c$ are in the interval $(-2,0)$.
As $2a^2+1\geq 2b^2+1$ and $2a^2 +1 \geq 2c^2+1$ we get by multiplying with $\frac{2b^2+1}{2a^2+1} \leq 1$ and $\frac {2c^2+1}{2a^2+1} \leq 1$ that
$\frac{2b^2+4b}{2b^2+1} \geq \frac{2b^2+4b}{2b^2+1}\cdot\frac{2b^2+1}{2a^2+1}=\frac{2b^2+4b}{2a^2+1}$ etc.
Reducing to a common denominator and applying $a+b+c=0$ we get the inequality.
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0I'll rethink everything again, thanks for your input though, it's highly appreciated :) – 2012-09-21