Let $a, b, c \in (0,1)$ and $a + b + c + a b + b c + c a = 1 + a b c$. Prove that:
$\displaystyle\frac{1+a}{1+a^2} + \frac{1+b}{1+b^2} + \frac{1+c}{1+c^2} \leq \frac{3}{4}(3+\sqrt{3})$.
Let $a, b, c \in (0,1)$ and $a + b + c + a b + b c + c a = 1 + a b c$. Prove that:
$\displaystyle\frac{1+a}{1+a^2} + \frac{1+b}{1+b^2} + \frac{1+c}{1+c^2} \leq \frac{3}{4}(3+\sqrt{3})$.
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
If $f(x)=\sin x,f'(x)=\cos x, f''(x)=-\sin x<0$ as $\frac{\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
If $f(y)=\cos y,f'(y)=-\sin y, f''(y)=-\cos y<0$ as $-\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.