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I don't know how to prove that the following inequality holds (under condition $a+b+c=0$): $$\frac{(2a+1)^2}{2a^2+1}+\frac{(2b+1)^2}{2b^2+1}+\frac{(2c+1)^2}{2c^2+1}\geqq 3$$

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Substitution of $c=-a-b$ gives us a minimization problem in two unconstrained variables. As a rational expression, this function of $a$ and $b$ will have finitely many points critical to both partial derivatives. If the inequality holds on those finite number of points (here, 22 ordered pairs, with some symmetry), the inequality holds in general. As a disclaimer, this is an ugly approach. –  A Walker Sep 21 '12 at 15:31
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What have you tried? What reason do you believe it is true? (e.g. Is it an exercise from somewhere?) –  Thomas Andrews Sep 21 '12 at 15:39
    
I suspect you need to prove that the product of these terms is greater than one, and hence the arithmetic mean is greater than one, and hence that their sum is greater than 3. –  Thomas Andrews Sep 21 '12 at 16:04
    
I 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. –  Tunger Sep 21 '12 at 17:58

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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 $$\sum_\text{cyc}\frac{(2a+1)^2}{2a^2+1}\geq3\sum_\text{cyc}\frac{(2a+1)^2}{4(a^2+b^2+c^2)+3}=3\left(1+\frac{4(a+b+c)}{4(a^2+b^2+c^2)+3}\right)=3.$$

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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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Just because $a>b>c$ doesn't mean that $2a^2+1>2c^2+1$ ($c$ is negative.) You really want $|a|\geq|b|\geq|c|$ –  Thomas Andrews Sep 21 '12 at 15:41
    
I hope it's fixed now, thanks for pointing out that mistake. –  Long Sep 21 '12 at 15:42
    
I'm not sure where the $3\cdot ...$ part comes into the equation. –  Thomas Andrews Sep 21 '12 at 15:44
    
Finally, is it true that if $2a^2+1>2b^2+1$ that necessarily $$\frac{2b^2+4b}{2b^2+1}\geq \frac{2b^2+4b}{2a^2+1}$$? What if $2b^2+4b<0$? –  Thomas Andrews Sep 21 '12 at 15:47
    
I'll rethink everything again, thanks for your input though, it's highly appreciated :) –  Long Sep 21 '12 at 15:50

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