It is given that $$P_x = \frac{\int_0^x w^{{(n-2) \over 2}} \exp(-{w \over 2})\ dw}{\Gamma{({n \over 2})}\ 2^{{n \over 2}}}, \text{ for } x>0$$
My approach:
$$P_x = 1 - \frac{\int_x^\infty w^{{(n-2) \over 2}} \exp(-{w \over 2})\ dw}{\Gamma{({n \over 2})}\ 2^{{n \over 2}}}$$
$$1 - P_x = \frac{\int_x^\infty w^{{(n-2) \over 2}} \exp(-{w \over 2})\ dw}{\Gamma{({n \over 2})}\ 2^{{n \over 2}}}$$
$$\frac{1 - P_x}{n} = \frac{\int_x^\infty w^{{(n-2) \over 2}} \exp(-{w \over 2})\ dw}{\Gamma{(1 +{n \over 2})}\ 2^{1 + {n \over 2}}}$$
How should I proceed from here? I have to show that $x < {n \over 1-P_x}$.
This is a simple application of Markov's inequality which states that for a non-negative random variable $X$ and $x>0$, $$\dfrac{\mathbb{E}(X)}x\ge \mathbb{P}(X\ge x)$$
Choosing $X\sim\Gamma\left(\frac{n}2,\frac12\right)$, where the parameters are shape and rate parameters respectively, $\mathbb{E}(X)=n, \mathbb{P}(X\ge x)=1-P_x$ since $X$ is a continuous random variable, we get the inequality you desire.