I5-6600K Vcore AsRockExtreme4

Do you have speedstep enabled? Try setting Windows Power Management to balanced.

The CPU Vcore Voltage increases at idle and decreases under load (to reduce the power because the current is higher under load). Why would you want yours to be any different? This is the normal way of these things.

It reduces your CPU frequency, not speed.

then... what is cpu speed?

As I wrote, the clock rate in Hertz (or cycles per second as you put it) is the frequency of the CPU, not its speed. This is why a CPU with a higher clock rate than another can be slower (i.e. have less speed) than one with a lower clock rate. You just have to benchmark this to prove it again and again and the reason is that the speed of a CPU is function of a lot more parameters than solely its clock rate e.g. its memory quantity and clock rate, its cache size, its architecture (like comparing Skylake with Ivy Bridge), etc... You still have a long way to go. Good luck with your overclocking ;-).

Well you have to measure it. That's why all those benchmarks are around, and why there are all different. You can't compare 867 on Cinebench R15 (excellent) with 7600 (poor) on PassMark. For a car you just have to give a number of Km (or Miles) per hour and speed is clear. It is certainly not that simple for a CPU because there we refer to the speed "of execution", not of displacement which is much simpler (assuming the road is kept flat and dry, the wind is the same, etc...).

The speed of execution of a CPU depends on a number of factors such as the type load of the CPU (to name but one) - for example the speed of execution will vary considerably between arithmetic calculation (like Prime95 Blend test) vs. image rendering (like Cinebench R15 CPU test) because this relates to the speed of execution of its instructions in machine code, and not all the instructions are the same - some take fewer clock cycles than others. Multiple floating point calculations for example need a lot of "carry" micro code instructions which consume multiple cycles each and therefore can considerably slow down a CPU.

It's a bit like for measuring the amount of work a car can go through in one hour. Would you say that a (fast) Ferrari performs more work than a (slow) agricultural tractor? Yes, it's true on the freeway, but not in the field (or forest). This is where the car racer and the farmer would have very different (and divergent) opinions about the speed (i.e. the performance) of their respective vehicles.

Think about it...

On the contrary, my sentence makes perfect sense and although I admit that it may be seen as being counter intuitive, it is absolutely correct (I propose to open the bet at USD1M cash - this is really easy money - I will soon be filthy rich then, can't wait) . Of course you have to understand this subject, have sufficient knowledge about what you are talking about, or alternatively know how to use the internet (things like Google anybody?) and find something that's plastered all over (nothing short of being common knowledge by now, especially after so many years).

Take a look here:
http://www.tweaktown.com/guides/7481/tweaktowns-ultimate-intel-skylake-overclocking-guide/index6.html
This is "TweakTown's Ultimate Intel Skylake Overclocking" and the graph under "Serial Voltage Identification" clearly shows how CONSISTENTLY LOWER the CPU Auto VCore (SVID) is under load, particularly at 4.8 GHz where it FALLS from 1.325 V all the way To A MUCH LOWER 1.284 V (!) and they mention under the graph "You can see the large drop from idle to load voltages." And these are not completely stupid people either (they do measure the voltage on the CPU pins directly with an electronic multimeter - no fancy CPU-Z action, smoke and mirrors).

Of course I have written about this more than once in other posts, but regardless, it's plastered all over the internet. Just look here:
https://pcpartpicker.com/forums/topic/130888-overclocking-i5-6600k Very first comment at the top: "That's completely normal. Its caused by a thing called Vdroop. When your CPU goes under load it causes the Vcore to drop."... too easy,

As I wrote previously, [Power P (W) = Voltage (V) x Current (A)] as in [72 W = 12 V x 6 A] and since the current is higher under load (this is not counter-intuitive), the motherboard (not the CPU) reduces the voltage under load (as a self-protection mechanism to avoid damaging the CPU) by an amount equal to Vdroop (V). Mine goes form 1.356 V (IDLE) to 1.344 V (UNDER LOAD) - clearly a drop of 12 mV (0.012 V).

If you "want a stable temperature and a good CPU life at higher clocks than the factory ones," you'd better start learning (applying yourself) about topics like "CPU Load Line Calibration" (LLC) and how it reins in the effect of Vdroop. I run my i5 6600K at 4.6 GHz after manually setting CPU VCore at 1.355 V in the BIOS completely stable for hours and hours of testing with Prime95 both Blend and Heat tests. From this OC, I run my 16 GB (4 x 4 GB) HyperX Fury Black C14 2133 MHz memory chips at 3200 MHz under 1.30 V (from 1.20 V), and my two GTX 970 graphics cards in SLI with their GDDR5 memory at over 4000 MHz (without changing the stock voltage) to cope with their heavily OCd GPUs. My case temps have been optimised with a laser guided infra-red thermal sensor to reach temps at or below 60 C max. under Prime95 Blend test with a budget single fan CPU cooler (no water in this fine electronics).

So, please don't tell me that I write a sentence that "doesn't make any sense". And concerning your "as far as I'm aware"... well, let me tell you that you are "obviously" far from being aware enough, and therefore you still have a long... very long way to go.

Ah!...Google... if I remember well (from 20 seconds ago) that's the name of a search engine on the internet... I suggest that next time, you check it out before you "expose naked" your complete UN-awareness, or better still, learn how to use it before you undertake a task such as the overclocking of a modern CPU. Good luck with your OC. ;-)

^ chin up