a,b,c>0 and a^4+b^4+c^4 = 3 . Prove the :
sigma\frac{1}{4-ab} =< 1
Do you mean this?Originally Posted by math
If $\displaystyle a, \, b, \, c > 0$ and $\displaystyle a^4 + b^4 + c^4=3$, then $\displaystyle \sum \frac{1}{4-ab} \le 1$
What kind of sum? There is only one term. Or do you mean
$\displaystyle \frac{1}{4-ab} + \frac{1}{4-bc} + \frac{1}{4-ac}\le 1$?
This inequality appears to be correct. It can probably be proved using Lagrange multipliers: Set $\displaystyle F(a,b,c) = a^4 + b^4 + c^4$ and $\displaystyle G(a,b,c) = \frac{1}{4-ab} + \frac{1}{4-bc} + \frac{1}{4-ac}$. You want to find the maximum of G subject to F(a,b,c) = 3. The Lagrange equations are
$\displaystyle \frac{b}{(4-ab)^2} + \frac{c}{(4-ac)^2} = 4 \lambda a^3$
$\displaystyle \frac{a}{(4-ab)^2} + \frac{c}{(4-bc)^2} = 4 \lambda b^3$
$\displaystyle \frac{b}{(4-bc)^2} + \frac{a}{(4-ac)^2} = 4 \lambda c^3$
Good luck