Is this inequality true? Prove or Disprove it!

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The inequality (A+B)^{p} ≤ p(A^{p}+B^{p}) is under scrutiny for its validity, particularly for irrational values of p. Initial attempts to prove it using the binomial theorem were unsuccessful, suggesting the need for additional conditions. Testing specific values, such as A=0, indicates that p must be greater than or equal to 1 for the inequality to hold. Further examples demonstrate that without restrictions on A, B, and p, the inequality does not consistently apply. Clarifying these parameters is essential for a proper proof or disproof of the inequality.
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I've encountered this nice-looking inequality:

<br /> \left(A+B\right)^{p} \le p\left(A^{p}+B^{p}\right)<br />

(p can irrational as well)

but I can't find a way to prove or disprove its correctness. I've tried using the binomial theorem, but it didn't seem it would lead me to the finish.

Could someone please tell me how to prove that?

Thank you very much!
 
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Yeah, looks nice. But there must be other premises. Taking A=0 shows you need p>=1. e.g.
 
What other info do you have?
For example, (2+3)^3 <= 3(2^3 + 3^3)
Nope, doesn't work.

(0 + 1)^(power) isn't going to be less than that power*(0^power + 1^power)

So, there must be some restriction on A, B, and p that you haven't stated.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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