0
$\begingroup$

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

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

  • 0
    o$f$ course, it 's ok2012-09-11

1 Answers 1

0

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.

  • 1
    @lab: May I direct your attention to [$\TeX$/$\LaTeX$/MathJax basic tutorial and quick reference](http://meta.math.stackexchange.com/q/5020), especially point 2. on displayed equations and the answer on [aligned](http://meta.math.stackexchange.com/a/5024) equations? This might improve readability of your answers which often feature somewhat complicated expressions in inline mode.2012-09-11