How to prove $\sum_{i=1}^na_i> n- \frac{9}{4}$ Given that $a_1=\frac{1}{2}$ and $a_{n+1}=\sqrt[]{\frac{1}{3}a_n^{2}+\frac{2}{3}a_n}$ ,prove that
$$\sum_{i=1}^na_i> n- \frac{9}{4}$$Thanks.
 A: Let $f(x)=\sqrt{\frac{1}{3}(x^2+2x)}$, for $x\ge0$. Clearly $f$ is increasing and it is staightforward to check that 
$$0\le x\le 1\Longrightarrow 0\le f(x)\le x\le 1$$
This alows us to prove that the sequence $(a_n)$ is increasing and bounded by $1$, so it must converge to the positive solution of the equation $f(x)=x$ which is $1$. Moreover, $\frac12\le a_n\le1$ for every $n$.
Now, 
$$f'(x)=\frac{1}{\sqrt{3}}\frac{x+1}{\sqrt{x^2+2x}}=\frac{1}{\sqrt{3}}\sqrt{1+\frac{1}{x^2+2x}}$$
So, $f'$ is decreasing on $[\frac12,1]$. So, 
$$0\le f'(x)\le f'(1/2)=\sqrt{3/5}\quad\text{for $x\in[0.5,1]$.}$$
The mean value theorem allows us to write
$$0\le 1-f(x)\le (1-x) \sqrt{3/5}\quad\text{for $x\in[0.5,1]$}$$
So, if $\delta_n=1-a_n$, then we have, for every $n$,
$$\delta_{n+1}\le \sqrt{3/5}\delta_n$$
Consequently,
$$\delta_n\le\left(\frac35\right)^{n/2}\delta_0$$
Thus
$$\sum_{k=1}^n\delta_k<\frac12\sum_{k=1}^\infty\left(\frac35\right)^{k/2}=
\frac{\sqrt3}{2(\sqrt5-\sqrt3)}\approx 1.72<\frac{9}{4}$$
This is equivalent to the desired inequality.
