If a,b,c>0 and a+b+c=3 prove that

$\displaystyle \frac{1}{a^{2}-a+3}+\frac{1}{b^{2}-b+3}+\frac{1}{c^{2}-c+3}\leq 1$

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- Aug 6th 2010, 09:34 PMbanku12Inequality 3
If a,b,c>0 and a+b+c=3 prove that

$\displaystyle \frac{1}{a^{2}-a+3}+\frac{1}{b^{2}-b+3}+\frac{1}{c^{2}-c+3}\leq 1$ - Aug 16th 2010, 08:40 AMthorin
Hi,

firstly, we can notice that $\displaystyle f(x)=\frac{1}{x^2-x+3}$ is concave on $\displaystyle [0,\frac{3}{2}]$ (by 2 derivations). So, if a,b, and c are in $\displaystyle [0,\frac{3}{2}]$, we can apply the inegality of convexity, and the result is obvious.

Now, if at least one of the 3 numbers is $\displaystyle > \frac{3}{2}$. Then, the 2 others are $\displaystyle < \frac{3}{2}$. Let's say that $\displaystyle c> \frac{3}{2}$ and $\displaystyle a< \frac{3}{2}$ and $\displaystyle b< \frac{3}{2}$, for example.

Then $\displaystyle f(c) \leq f(\frac{3}{2})$ and $\displaystyle f(a) \leq f(\frac{1}{2})$ and $\displaystyle f(b) \leq f(\frac{1}{2})$(because $\displaystyle f$ has a maxima in $\displaystyle x=\frac{1}{2}$ and is decreasing on$\displaystyle [\frac{1}{2},3]$.

finally, $\displaystyle f(a)+f(b)+f(c) \leq 2f(\frac{1}{2})+f(\frac{3}{2})<1$

is there a mistake ? - Aug 16th 2010, 10:13 AMAlso sprach Zarathustra
I love your solution!