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Prove: Let $p$ be an integer greater than $1$. Suppose $u,v$ are positive real numbers. Then $\left(\frac{u+v}{2}\right)^p\leq\frac{u^p+v^p}{2}$.

I only know $(u+v)^p\geq u^p+v^p$. How to continue the proof? Or I should start in another way?

I also think inversely, and I will get $(u+v)^p\leq 2^{p-1}(u^p+v^p)$, am I correct? But how to get this? Can somebody teach me pls, thx

Martin Sleziak
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sunny
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  • Do you know convexity? – nejimban Mar 20 '22 at 05:17
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    This follows by convexity of $x^{p}$ and it has been answered many times on MSE. – Kavi Rama Murthy Mar 20 '22 at 05:18
  • @nejimban sorry..I haven't learnt that, is there any other ways to solve it? :( – sunny Mar 20 '22 at 05:19
  • @KaviRamaMurthy Sorry, I haven't learnt convexity...My professor won't allow me to use concepts that haven't been taught in class to solve it:( – sunny Mar 20 '22 at 05:21
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    Minimize $2^{p-1}(1+t^{p})-(1+t)^{p}$ over $[1,\infty)$ and put $t=\frac u v$ in the resulting inequality. – Kavi Rama Murthy Mar 20 '22 at 05:26
  • Note that the duplicate still applies as long as one-variable calculus knowledge is assumed, since it involves only a maxima-minima problem and a usage of the modulus of a complex number, which one is likely to be familiar with if one is familiar with one-dimensional calculus (and is not required here, since $u,v$ are positive real numbers so we can replace $|\cdot|$ with $|\cdot|$) – Sarvesh Ravichandran Iyer Mar 21 '22 at 08:07

1 Answers1

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Hints: apply binomial theorem like this:

$$a^p+b^p=[\frac 12(a+b)+\frac 12(a-b)]^p+[\frac 12(a+b)-\frac 12(a-b)]^p$$

After expansion and simplification we have:

$$\frac 12 (a^p+b^p)=\big(\frac{a+b}2\big)^p+p\big(\frac{a+b}2\big)^{p-1}\big(\frac{a-b}2\big)+\frac{p(p-1)}{2!}\big(\frac{a+b}2\big)^{p-2}\big(\frac{a-b}2\big)^2+\cdot \cdot\cdot$$

If p is a positive integer or, any negative quantity(integer or fraction), then all terms on RHS are positive and hence we have:

$$\frac{a^p+b^p}2>\big(\frac{a+b}2\big)^p$$

Equality is when $a=b$

Comment:

Referring your question in comment, those are AM-GM inequalities and their proof are different. You can ask it in another question.

If your happy with my answer do not forget the tick for accepting the answer.

sirous
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  • Thank you! I'd like to ask can I also apply this method to ((a+b+c+d)/4)^p less than or equal to (a^p+b^p+c^p+d^p)/4, where a,b,c,d are positive real numbers? and also ((a+b+c)/3)^p less than or equal to (a^p+b^p+c^p)/3 ? Are these cases similar? – sunny Mar 20 '22 at 08:00