Polynomial Centuries

$P(x)=x^{100}+a_{99}x^{99}+a_{98}x^{98}+...+a_{2}x^2+a_{1}x+1$

It is given that all the roots of the equation $P(x)=0$ are real, meaning that there are 100 real roots. Since all coefficients of the polynomial $P(x)$ are real and positive, all the 100 roots of the equation $P(x)=0$ must be negative. Let these roots be $(-\alpha_i)$ , where $\alpha_i>0\;\forall\;i\;\in \{1,2,3,...,100\}$ . Therefore,

$P(x)=(x+\alpha_{1})(x+\alpha_{2})(x+\alpha_{3})...(x+\alpha_{99})(x+\alpha_{100})$

$\Rightarrow P(1)=(1+\alpha_{1})(1+\alpha_{2})(1+\alpha_{3})...(1+\alpha_{99})(1+\alpha_{100})$

Now, using AM-GM Inequality, $(1+\alpha_i) \geq 2\sqrt{\alpha_i}\;\forall\;i\;\in \{1,2,3,...,100\}$ .

Therefore, $P(1) \geq 2^{100}\sqrt{\alpha_{1}\alpha_{2}...\alpha_{100}}$

Now, using Vieta's Formula, $\alpha_{1}\alpha_{2}...\alpha_{100}=1$ . Hence,

$P(1) \geq 2^{100}$

$\Rightarrow 1+a_{99}+a_{98}+...+a_{2}+a_{1}+1 \geq 2^{100}$

$\Rightarrow \displaystyle \sum_{i=1}^{99} a_{i} \geq 2^{100}-2=2(2^{99}-1)$

Therefore, $N=99$ .