Show that a statement is true by induction for $x,y > 0$ and $n \in N$.

Question

I ve been assigned the following problem:

How can we prove by induction that for every $x,y > 0$ and $n \in N$ it is true that: $$\frac{x^n+y^n}{2} \geq \bigg(\frac{x+y}{2}\bigg)^n$$

I have established a base of $n=1$ so that $\frac{x+y}{2} = \frac{x+y}{2}$ and have assumed that the statement is true for any natural $k$ but am having problem proving that it is true for $k+1$.

Any help is appreciated!

Does this answer your question? How to show that $(a+b)^p\le 2^p (a^p+b^p)$. Some of the answers there deals with your problem, which is a stronger statement. — player3236, Dec 03 '20 at 19:58
This one is more appropriate: https://math.stackexchange.com/questions/409604/proof-of-the-inequality-xyn-leq-2n-1xnyn — player3236, Dec 03 '20 at 19:59

score 1 · Answer 1 · answered Dec 03 '20 at 20:02

Wlog, let $x+y=1$ and the inequality becomes $$ x^n+(1-x)^n\ge \frac{1}{2^{n-1}}. $$ Let $$ f(x)=x^n+(1-x)^n $$ and then $$ f'(x)=n(x^{n-1}-(1-x)^{n-1}),f''(x)=n(n-1)(x^{n-2}+(1-x)^{n-2}).$$ Setting $f'(x)=0$ gives $x=\frac12$ and $f''(\frac12)>0$. Therefore $f(x)$ attains a local minimum $\frac{1}{2^{n-1}}$ at $x=\frac12$. Note $f(0)=f(1)=1$ and thus $f(x)$ attain the global minimum $\frac{1}{2^{n-1}}$ at $x=\frac12$; Namely $$ x^n+(1-x)^n\ge \frac{1}{2^{n-1}} $$ and the equal sign "=" holds iff $x=y\frac12$.

Show that a statement is true by induction for $x,y > 0$ and $n \in N$.

1 Answers1