Binomial Theorem Inequality

Question

How do I proof the right side of the following inequality:

$(x^p+y^p)\leq(x+y)^p\leq2^{p-1}(x^p+y^p)$ for $0\leq x,y$ and $p > 1$

The left side can be easily proved with the binomial theorem and I assume that the right side can be proved with $\sum_{n=0}^{p}\binom{p}{n}=2^p$, but I can't get it working.

Welcome to Mathematics Stack Exchange. Did you mean binomial where you typed binomical? — J. W. Tanner, Feb 02 '21 at 11:24
Right hand inequality follows from convexity of the function $x \to x^{p}$. — Kavi Rama Murthy, Feb 02 '21 at 11:36

score 0 · Answer 1 · answered Feb 02 '21 at 11:37

Hint

Neither inequality is true unless you place some restrictions on the values of $\ x\ $, $\ y\ $ and $\ p\ $. However, the inequality $$ \left(x+y\right)^p\le2^{p-1}\left(x^p+y^p\right) $$ is equivalent to $$ \left(\frac{x}{2}+\frac{y}{2}\right)^p\le\frac{x^p+y^p}{2}\ , $$ and this will hold if the function $\ x\mapsto x^p\ $ is convex on its domain of definition. I'll leave it to you to determine the values of $\ p\ $ for which this is the case.

score 0 · Accepted Answer · answered Feb 02 '21 at 11:41

RHS is mean power inequality:If $p>1$, then $$M_p>M_1$$

$$\implies \left(\frac{x^p+y^p}{2}\right)^{1/p} >\frac{x+y}{2}~~~(1)$$ $$\implies \left(\frac{x^p+y^p}{2}\right) >\left(\frac{x+y}{2}\right)^p~~~(2)$$ Also (2) is Jensen's inequality for $f(x)=x^p \implies f'(x)=px^{p-1}, f''(x)= p(p-1)x^{p-2}>0, ~if~ p>1.$$

Binomial Theorem Inequality

2 Answers2