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Let $a,b,$ and $c$ be positive real numbers with $ab+bc+ca = 1$. Prove that $$\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{3}{2}$$

Attempt

The $ab+bc+ca = 1$ condition reminds of the rearrangement inequality. Thus, I would say that $a^2+b^2+c^2 \geq ab+bc+ca = 1$ then rewrite the given inequality as $4(a+b+c)^2 = 4(a^2+b^2+c^2) + 8(ab+bc+ca) = 4(a^2+b^2+c^2) \leq 9(a^2+1)(b^2+1)(c^2+1)$

I don't know what to do next.

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By AM-GM $\sum\limits_{cyc}\frac{a}{\sqrt{a^2+1}}=\sum\limits_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq\frac{1}{2}\sum\limits_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{3}{2}$.

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  • $\begingroup$ What is the meaning of the symbol $\sum_{cyc}$? $\endgroup$ – MathOverview Jan 7 '16 at 18:11
  • $\begingroup$ Also, how did you use AM-GM to get $\sum \limits_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}} \leq \frac{1}{2} \sum \limits_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)=\frac{3}{2}$.? $\endgroup$ – user19405892 Jan 7 '16 at 18:11
  • $\begingroup$ @ MathOverview, for example for three variables: $\sum\limits_{cyc}a=a+b+c$, $\sum\limits_{cyc}\frac{a}{b}=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ $\endgroup$ – Michael Rozenberg Jan 7 '16 at 18:16
  • $\begingroup$ @user19405892, I used $xy\leq\frac{1}{2}(x^2+y^2)$. $\sum\limits_{cyc}\sqrt{\frac{a^2}{(a+b)(a+c)}}=\sum\limits_{cyc}\left(\sqrt{\frac{a}{a+b}}\cdot\sqrt{\frac{a}{a+c}}\right)\leq\frac{1}{2}\sum\limits_{cyc}\left(\frac{a}{a+b}+\frac{a}{a+c}\right)$. $\endgroup$ – Michael Rozenberg Jan 7 '16 at 18:21
  • $\begingroup$ Why is $\sum\limits_{cyc}\frac{a}{\sqrt{a^2+1}}=\sum\limits_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}$? $\endgroup$ – user19405892 Jan 8 '16 at 1:39
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Set $a=\cot A$ etc.

$\implies\sum\cot A\cot B=1$

$\iff\tan A+\tan B+\tan C=\tan A\tan B\tan C$

$\implies A+B+C=n\pi$ where $n$ is any integer

Now use In $ \triangle ABC$ show that $ 1 \lt \cos A + \cos B + \cos C \le \frac 32$

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See my previous answer here: How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{yz}{x^2+2016}}\le\frac{3}{2}$

For an elegant solution, apply the so-called Purkiss Principle.

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  • $\begingroup$ Hi James. A general guideline to follow here on MathSE is that an answer should be self-contained. It would be a good idea to include the statement of the Purkiss Principle in your answer (and maybe a direct link to a proof or explanation of it if you think it would help the reader). And also your response should reference this questions directly, rather than just say, "go see this other answer elsewhere." $\endgroup$ – Mike Pierce Oct 17 '17 at 15:51

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