Inequality about the sum of square root of 1 over x This question originates from an adversarial bandit problem. In the proof of self-bounding property of regret, there is a step showing 
$$C\sum_{t=1}^T\sum_{i\neq i^*}\sqrt{\frac{x_{i,t}}{t}}\leq 2C\sqrt{KT}$$
where $\sum_{i\neq i^*}$ means summing over all $K$ arms except the optimal arm $i^*$, and $x_{i,t}$ is a probability distribution over $K$ arms at time $t$ with $\sum_{i=1}^K x_{i,t}=1,\forall t$.
Now the hint in the notes says

this follows trivially by Jensen on the $i$ sum

I don't really get how to use Jensen's inequality to prove it, but what I got so far is to use Cauchy–Schwarz inequality to prove 
$$\sum_{i\neq i^*}\sqrt{x_{i,t}}\leq \sqrt{K},\forall t$$
which lefts me with
$$\sum_{t=1}^T \sqrt{\frac{1}{t}}\leq 2\sqrt{T}$$
to prove. The way I got around it is to let $f(T) = 2\sqrt{T}-\sum_{t=1}^T \sqrt{\frac{1}{t}}$, first show that $f(1)=0$, and then show that $f'(T)\geq 0$ for $T>0$ (help from Wolfram Alpha) which verifies $f(T)$ is monotonically increasing for positive $T$. However, this approach involves dealing with HurwitzZeta function, which is a bit complicated.
I wonder if there is some handy tricks I can use to prove the inequality about $f(T)$, or if there is some more direct way to use Jensen's inequality to prove the overall inequality, as the hint suggests.

The best way I now know to see the inequality about $t$ is to upbound the sum by integral:
$$\sum_{t=1}^T\sqrt{\frac{1}{t}}\leq \int_0^T\sqrt{\frac{1}{t}} dt= 2\sqrt{T}$$
Of course, the upboundedness is true because $\sqrt{\frac{1}{t}}$ is monotonically decreasing. 
A lesson I took from it was while we can approximate integral by (Riemann) sum, we can also approximate/bound sum by integral. This approximation trick can be quite useful given conditions like monotonicity. 
