$1/(51 +u^2) + 1/(51 +v^2) + 1/(51 +w^2)$ inequality I need to prove that $$1/(51 +u^2) + 1/(51 +v^2) + 1/(51 +w^2) \leq 1/20$$ given $u + v + w = 9$ and $u,v,w$ positive reals.
Using AM-HM inequality with $(51 +u^2, 51 +v^2, 51 +w^2)$ I arrive at 
$(51 +u^2 + 51 +v^2 + 51 +w^2)/3 \geq 3/((1/(51 +u^2) +1/(51 +v^2) 1/(51 +w^2))$, $$or,$$ 
$1/(51 +u^2) +1/(51 +v^2)+1/(51 +w^2)  \leq 9/(51 +u^2 + 51 +v^2 + 51 +w^2)$
In order to take the next step, I need to estimate the RHS with an upper bound. Here is where I face the problem. With $(u+v+w)^2 \leq 3(u^2 + v^2 + w^2)$ from  C-S inequality and $u+v+w =9$ we get  $27 \leq  (u^2 + v^2 + w^2)$ which does not provide an upper bound. I tried several other inequalities, but each lead to a bound on the "wrong" side. Am I at all looking in the right direction? 

You're doing well. Note that you have $u^2+v^2+w^2$ in the denominator, and $1/x$ is a decreasing function. That means that
$$ \left( u^2+v^2+w^2 \ge \frac{(u+v+w)^2}{3} \right) \Rightarrow \left( \frac{9}{153+u^2+v^2+w^2} \le \frac{9}{153+\frac{(u+v+w)^2}{3}} \right)$$
EDIT: The answer is not correct, because the inequality $$\frac{1}{51+u^2} +\frac{1}{51+v^2}+ \frac{1}{51+w^2} \le \frac{9}{153+u^2+v^2+w^2} $$ is not correct. The correct one has $\ge$ sign, which doesn't help.
