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Let $a, b, c \in (0,1)$ and $a + b + c + a b + b c + c a = 1 + a b c$. Prove that:

$\displaystyle\frac{1+a}{1+a^2} + \frac{1+b}{1+b^2} + \frac{1+c}{1+c^2} \leq \frac{3}{4}(3+\sqrt{3})$.

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$\frac{1+a}{1+a^2}+\frac{1+b}{1+b^2}+\frac{1+c}{1+c^2}$ – LevanDokite Sep 11 '12 at 13:46
Perhaps the notation $\sum_{cyc}$ for cyclic sum could be also appropriate in this case, but $\frac{1+a}{1+a^2}+\frac{1+b}{1+b^2}+\frac{1+c}{1+c^2}$ looks better, in my opinion. – Martin Sleziak Sep 11 '12 at 13:59
Since you accepted the edit, I removed my comments. – Rod Carvalho Sep 11 '12 at 14:00
of course, it 's ok – LevanDokite Sep 11 '12 at 14:35
up vote 0 down vote accepted

Put $a=\tan A,b=\tan B, c=\tan C$

So, $\tan(A+B+C)=1\implies A+B+C=\frac{\pi}{4} $ .

Now, $$\frac{1+a}{1+a^2}=\frac{1+\cos2A+\sin2A}{2}=\frac{1+\sqrt2\sin(2A+\frac{\pi}{4})}{2}$$

If $x=2A+\frac{\pi}{4}$, as $0<A<\frac{\pi}{4}\implies \frac{\pi}{4}<x<\frac{3\pi}{4}$

If $f(x)=\sin x,f'(x)=\cos x, f''(x)=-\sin x<0$ as $\frac{\pi}{4}<x<\frac{3\pi}{4}$

So, $\sin x$ is concave function in $(\frac{\pi}{4},\frac{3\pi}{4})$.

So using Jensen's inequality,

$$\sum \sin(2A+\frac{\pi}{4})≤3\sin\left(\frac{2A+\frac{\pi}{4}+2B+\frac{\pi}{4}+2C+\frac{\pi}{4}}{3}\right)$$ $$=3\sin\frac{5\pi}{12}=3\sin(\frac{\pi}{4}+\frac{\pi}{6})=\frac{3(\sqrt3+1)}{2\sqrt2}$$

So, $$\sum\frac{1+a}{1+a^2}=\sum\frac{1+\sqrt2\sin(2A+\frac{\pi}{4})}{2}$$ $$=\frac{3}{2}+\frac{1}{\sqrt2}\sin(2A+\frac{\pi}{4})$$ $$≤\frac{3}{2}+\frac{1}{\sqrt2}\frac{3(\sqrt3+1)}{2\sqrt2}$$ $$=\frac{3}{2}+\frac{3(\sqrt3+1)}{4}=\frac{3\sqrt3}{4}+\frac{9}{4}$$

Alternatively, $$\frac{1+a}{1+a^2}=\frac{1+\cos2A+\sin2A}{2}=\frac{1+\sqrt2\cos(2A-\frac{\pi}{4})}{2}$$

If $y=2A-\frac{\pi}{4}$, as $0<A<\frac{\pi}{4}\implies -\frac{\pi}{4}<y<\frac{\pi}{4}$

If $f(y)=\cos y,f'(y)=-\sin y, f''(y)=-\cos y<0$ as $-\frac{\pi}{4}<y<-\frac{\pi}{4}$

So, $\cos y$ is concave function in $(-\frac{\pi}{4},\frac{\pi}{4})$.

So using Jensen's inequality,

$$\sum \cos(2A-\frac{\pi}{4})≤3\cos\left(\frac{2A-\frac{\pi}{4}+2B-\frac{\pi}{4}+2C-\frac{\pi}{4}}{3}\right)$$ $$=3\cos(-\frac{\pi}{12})=3\cos(\frac{\pi}{6}-\frac{\pi}{4})=\frac{3(\sqrt3+1)}{2\sqrt2}$$ and so on.

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but how can we know ? $$tanA+tanB+tanC+tanA.tanB+tanB.tanC+tanC.tanA=tanA.tanB.tanc+1$$ and $tan(A+B+C)=\pi$ – LevanDokite Sep 11 '12 at 13:23
$\frac{a+b+c-abc}{1-ab-bc-ca}=1$, right? – lab bhattacharjee Sep 11 '12 at 13:44
ok i understand – LevanDokite Sep 11 '12 at 13:47
@lab: May I direct your attention to $\TeX$/$\LaTeX$/MathJax basic tutorial and quick reference, especially point 2. on displayed equations and the answer on aligned equations? This might improve readability of your answers which often feature somewhat complicated expressions in inline mode. – t.b. Sep 11 '12 at 14:11

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