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Hy all!

I’m having trouble finding a proof for the following problem:

Show that $a^x+b^x+c^x>(a+b+c)^x$, if $a,b,c>0$ and $0<x<1$ (over the real numbers).

This inequality has been torturing me for long hours the least. I couldn’t find anything on Google. This kind of equation isn’t made for search engines tbh.

- $2^n=C_0+C_1+\dots+C_n$
- Inverse of $y = x^3 + x $?
- Solving $x^3 + x^2 - 4 = 0$
- Range of a 1-2 function
- What does this equal? $6\div 2(1+2)$
- Polar form of the sum of complex numbers $\operatorname{cis} 75 + \operatorname{cis} 83 + \ldots+ \operatorname{cis} 147$

Anyway, I really thinked about it a lot but didn’t make any significant progress.

It would be great to see a proof without the really high-end Mathematics.

Any ideas to start with?

- absolute minimum of function
- Prove triangle inequality using the properties of absolute value
- Polar equation of a circle
- Is there a finite number of $(a, b)$ pairs that satisfy, ${a+b} \leq \frac {{a^2}b+{b^2}a}{a^2 +b^2}$
- An inequality, which is supposed to be simple
- Proof that Right hand and Left hand derivatives always exist for convex functions.
- $2^n=C_0+C_1+\dots+C_n$
- Will inverse functions, and functions always meet at the line $y=x$?
- Why is the even root of a number always positive?
- How prove this inequality $3a^3b+3ab^3+18a^2b+18ab^2+12a^3+12b^3+40a^2+40b^2+64ab\ge 0$

First, prove the inequality for two terms $a,b$. I essentially repeat the proof from Prove that $(p+q)^m \leq p^m+q^m$: let $y=1-x$ and

$$(a+b)^x = (a+b)^{1-y} = a (a+b)^{-y} + b (a+b)^{-y} < a a^{-y} + b b^{-y} = a^x + b^x$$

Now you can the inequality for three or more terms just by using parentheses:

$$(a+(b+c))^x < a^x+(b+c)^x < a^x+b^x+c^x$$ (This CW answer is a compilation of the comments above )

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