# How to show that $(a+b)^p\le 2^p (a^p+b^p)$ [duplicate]

If I may ask, how can we derive that $$(a+b)^p\le 2^p (a^p+b^p)$$ where $a,b,p\ge 0$ is an integer?

• If $p$ is odd, you also need $a,b \ge 0$. Jul 20, 2014 at 8:07

Assuming $a,b,p\geq 0$

$$(a+b)^p\leq\left(2\cdot\max(a,b)\right)^p= 2^p\left[\max(a,b)\right]^p=2^p\max(a^p,b^p)\leq2^p\left(a^p+b^p\right)$$

• Is there a further proof which shows $$[\max(a,b)]^p=\max(a^p,b^p)$$ which you asserted along the way? Maybe it's trivial but that's the only equality I don't see 100% clearly yet. Jul 20, 2014 at 4:42
• The function $x^p$ is strictly increasing on $[0,\infty)$ for $p> 0$ and constant for $p=0$.
– user123641
Jul 20, 2014 at 4:44
• @glace : Just check the cases $a \le b$ and $b \le a$, assuming again $p > 0$. Jul 20, 2014 at 9:08

Hint: $$A^kB^{p-k}\leq A^p+B^p$$

In fact we can make the stronger claim:

$$(a+b)^p\le 2^{p-1} (a^p+b^p)$$

if $a,b \ge 0$. As Mohammad Khosravi points out, this is equivalent to

$$(x+1)^p\le 2^{p-1} (x^p+1)$$

if $x \ge 0$. We prove this by induction. The case $p=1$ is easy: $x+1 \le x+1$. So suppose we have established that

$$(x+1)^{p-1}\le 2^{p-2} (x^{p-1}+1)$$

Then we have:

\begin{align} (x+1)^p & = (x+1)(x+1)^{p-1}\\ & \le 2^{p-2}(x+1)(x^{p-1}+1) \end{align}

Now we just have to show that $2^{p-2}(x+1)(x^{p-1}+1) \le 2^{p-1} (x^p+1)$, or equivalently that $2(x^p+1) - (x+1)(x^{p-1}+1) \ge 0$:

\begin{align} 2(x^p+1) - (x+1)(x^{p-1}+1) & = x^p-x^{p-1}-x+1 \\ & = (x-1)(x^{p-1}-1) \\ & = (x-1)^2(x^{p-2} + x^{p-3} + \ldots + x + 1) \\ & \ge 0 \end{align}

Using Jensen's Inequality, we get for $p\ge1$ or $p\le0$, $$\left(\frac{A+B}2\right)^p\le\frac{A^p+B^p}2$$ Which, upon multiplication by $2^p$, yields $$(A+B)^p\le2^{p-1}(A^p+B^p)$$

Applying Jensen inequality on the convex function $f (x)=x^p$, $p> 1$

Assume that both $A, B > 0$ and $p \geq 0$ to begin with, then we can rewrite the inequality as:

$\left(\dfrac{A}{A+B}\right)^p + \left(\dfrac{B}{A+B}\right)^p \geq 2^{-p}$. So there are $3$ cases to consider:

1. $p = 0$, then $LHS = 2 > 1 = RHS$, and the inequality holds.

2. $p = 1$, then $LHS = 1 > \dfrac{1}{2} = RHS$, and the inequality also holds.

3. $p \geq 2$, then $f(x) = x^p$ is convex, so $f(x) + f(y) \geq f\left(\dfrac{x+y}{2}\right)$

with $x = \dfrac{A}{A+B}$, and $y = \dfrac{B}{A+B}$ yields the inequality.

• It seem we posted together, sorry! Jul 20, 2014 at 4:45
• it happens all the times here.... Jul 20, 2014 at 4:47

By induction.

Assume without loss of generalization that $A \leq B$

Note that $A^{n-1}(B-A) \leq B^{n-1}(B-A) \Rightarrow A^{n-1}B - A^n \leq B^n - AB^{n-1} \Rightarrow AB^{n-1} + A^{n-1}B) \leq A^n + B^n$

For $p = 1$ this is obvious. Supose true for all $p < n$.

$(A+B)^n = (A+B)^{n-1}(A+B) \leq 2^{n-1}(A^{n-1}+B^{n-1})(A+B) = 2^{n-1}(A^n + B^n + AB^{n-1} + A^{n-1}B) \leq 2^{n-1}(A^n + B^n + A^n + B^n) = 2^n(A^n + B^n)$

One other idea besides the given answers is like this:

if one of $a,b$ be zero, nothing remains for proof. Without loss of generality, let $a\ne 0$ and $x = a^{-1}b$, then one needs to prove that $$0 \le 2^p(x^p+1)-(x+1)^p =: f(x)$$ Now using $f,f',...$ one can derive a proof!