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Given three positive numbers a,b,c satisfying $a+b+c=3$. Show that $\sum\limits_{cyc}\frac{1}{(b+c)^2+a^2}\leq \frac{3}{5}$

Things I have done so far: $$a+b+c=3\Rightarrow b+c=3-a;0<a<3$$ $$\Rightarrow \frac{1}{(b+c)^2+a^2}=\frac{1}{(3-a)^2+a^2}=\frac{1}{2(a-1)^2+7-2a}\leq \frac{1}{7-2a}$$ Then, I tried to use the UCT to solve this problem. I created the new inequality: $$\frac{1}{7-2a}\leq \frac{1}{5}+m.(a-1)(*)$$ with $0<a<3$. I needed to find "m" which make (*) always true.After that, I found $$m=\frac{2}{25}$$ However, $$(*)\Leftrightarrow \frac{1}{7-2a}\leq \frac{1}{5}+\frac{2}{25}.(a-1)\Leftrightarrow 4(a-1)^{2}\leq 0$$ which is wrong with any $$a\in \mathbb{R} $$ Can you show me what my mistake is? I hope you can have "smart" way to solve this problem. Sorry, I am not good at English.

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    $\begingroup$ Add the source of the inequality (which competition, problem book, ..) and if possible, which methods you expect to need to prove it (e.g. if it appeared as an exercise in a section about Cauchy-Schwarz), or expected difficulty. $\endgroup$
    – Bart Michels
    Aug 7, 2018 at 11:54
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    $\begingroup$ @ShizumiAoki: this site is not really intended for posts of random problems like this, particularly if they have no source or motivation. In general, if all you can post is a problem statement, and your only source is "a friend", you are probably better off using a different website. $\endgroup$
    – Carl Mummert
    Aug 7, 2018 at 18:15
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    $\begingroup$ On the triangle given by $x+y+z=3$ and $x,y,z\geq 0$ the maximum of $\sum_{cyc}\frac{1}{x^2+(3-x)^2}$ can be found by Lagrange multipliers. $(1,1,1)$ is the only stationary point in the interior of the domain and the behaviour on the boundary is simple to study. Or you may prove that by replacing both $a$ and $b$ with their average the value of the function to maximize increases. $\endgroup$
    – Jack D'Aurizio
    Aug 7, 2018 at 18:15
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    $\begingroup$ Thank you for editing the question, Could you please also add the source of the problem and some motivation for why it is of interest? $\endgroup$
    – Carl Mummert
    Aug 8, 2018 at 0:50
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    $\begingroup$ @Carl Mummert is this restriction written down somewhere? The site tour includes "Solving mathematical puzzles" in the list of suitable questions. I have seen many interesting questions motivated only by idle curiosity, and well-received questions similar to this one (eg math.stackexchange.com/q/1857856). $\endgroup$
    – stewbasic
    Aug 8, 2018 at 4:59

2 Answers 2

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Hint: Use the Tangent Line method.

Indeed, we need to prove that $$\sum_{cyc}\frac{1}{(3-a)^2+a^2}\leq\frac{3}{5}$$ or $$\sum_{cyc}\left(\frac{1}{5}-\frac{1}{2a^2-6a+9}\right)\geq0$$ or $$\sum_{cyc}\frac{a^2-3a+2}{2a^2-6a+9}\geq0$$ or $$\sum_{cyc}\left(\frac{a^2-3a+2}{2a^2-6a+9}+\frac{1}{5}(a-1)\right)\geq0$$ or $$\sum_{cyc}\frac{(a-1)^2(2a+1)}{2a^2-6a+9}\geq0$$ and we are done!

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$$\sum_{cyc}\dfrac{1}{(b+c)^2+a^2} \le \dfrac{2}{25}(a+b+c)+\dfrac{9}{25}=\dfrac{3}{5}$$

$$\dfrac{1}{(b+c)^2+a^2}=\dfrac{1}{(3-a)^2+a^2}\le \dfrac{2}{25}a+\dfrac{3}{25}\ , (a\ge 0)$$

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