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It's easy to get the following estimate for a trinomial to the power of $2$, $$ (a + b + c)^2 \leq 3a^2 + 3b^2 + 3c^2. $$ Is there any analogous result for even powers, this is, for the more general case $(a + b + c)^{2n}$, with $n\in\mathbb{N}$?

Aguazz
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    Analogous could be ambiguous here. Does $(a+b+c)^{2n}\leqslant 3^{2n-1}(a^{2n}+b^{2n}+c^{2n})$ qualify? – Macavity Jun 21 '21 at 16:35
  • That's it!! Any proof? – Aguazz Jun 21 '21 at 16:45
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    Read up on Power Means Inequality or an extended version of Holder can also prove this. You already have an answer, a suitable definition of inner product also can prove it. – Macavity Jun 21 '21 at 16:47

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Depends on what you mean by "analogous", best I can think of is using the Cauchy–Schwarz inequality :

|<u,v>|² $\leq$ <u,u><v,v>

It applies to any pre-hilbert space , so you can produce "analogous" results by varying either the dimension of the space or the definition of inner product before appliying the inequality to well chosen vectors.

For example, take the euclidian space of dimension 3 with the usual definition of inner product. Let u=(x,y,z) and v=(1,1,1)

Applying the inequality gives : (x+y+z)² $\leq$ 3.(x²+y²+z²)

Other spaces will give you other inequalities.

Hope it helps :)

Anthony SPRIET
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    Hi Anthony!! Thanks for your answer :) By analogous, I mean something like $(a + b + c)^{2n} \leq k(a^{2n} + b^{2n} + c^{2n})$, for a constant $k$. – Aguazz Jun 21 '21 at 16:41
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    Then you should use Holder's Inequality, find more details here : https://math.stackexchange.com/questions/1070961/why-this-power-inequality-for-sums-of-real-numbers-holds

    To keep things short : $(x_1 + x_2 + ... + x_k)^{2n} \leq (k^{2n-1})(x_1^{2n} + x_2^{2n} + ... + x_k^{2n})$

     – Anthony SPRIET Jun 21 '21 at 17:22