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Given $a,b \geq 0$ and $2 \leq p < \infty$ prove that $(a^p + b^p)^{1/p} \leq (a^2 + b^2)^{1/2}$. I tried using the convex function $f(x) = x^p$ and then prove $a^p + b^p \leq (a^2 + b^2)^{p/2}$ but can't seem to make any real progress. Can someone help me?

nonuser
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Shuster
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    This is equivalent to showing $(1+ c^p)^{1/p} \leq (1+ c^2)^{1/2}$ or that $(1+ d^q)^{1/q} \leq 1+d$ for $q >1$ or that $1+ d^q \leq (1+d)^q$ for $q>1$ – Henry May 21 '21 at 22:04
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    related https://math.stackexchange.com/questions/2401054/if-n-geq-m-then-xmym1-m-ge-xnyn1-n – Albus Dumbledore May 22 '21 at 01:31

2 Answers2

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Let $c=a/b$ and observe $f:[2,\infty)\to \mathbb{R}$ defined by $$f(x) := (c^2+1)^x-(c^x+1)^2 $$then $$f'(x)=(c^2+1)^x\ln(c^2+1)-(c^x+1)c^x\ln c^2$$

If $f'(x)>0$ then $f$ is increasing so $f(x)\geq 0$ for all $x$ since $f(2)=0$

Clearly $\ln(c^2+1)>\ln(c^2)$ so it is enough to check if $$(c^2+1)^x \geq c^x(c^x+1)$$ for all $x\geq 2$, i.e. $$(c+{1\over c})^x \geq c^x+1$$

Let $$g(x) := (c+{1\over c})^x - c^x- 1$$ Then $$g'(x) = (c+{1\over c})^x\ln (c+{1\over c}) - c^x\ln c$$

Since clearly $g'(x)\geq 0$ we are done.

Thomas Andrews
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nonuser
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    Neat proof, thank you! – Shuster May 21 '21 at 22:12
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    It’s much easier to assume WLOG $a\leq b$ and thus $c\leq 1.$ Then $\log(c^2)\leq0$ and it follows directly that $$f’(x)=\text{positive}-\text{non-positive}>0,$$ without the rest of the proof. – Thomas Andrews May 22 '21 at 00:05
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A more general inequality is true ( probably already on this site, but could not find the duplicate)

If $a_i\ge 0$ and $0< x < y$ then $$\left( \sum a_i^y\right)^{\frac{1}{y}} \le \left( \sum a_i^x\right)^{\frac{1}{x}}$$ Indeed, consider $a_i^x= b_i$, and $z = \frac{y}{x}$. We have to show that $$\left(\sum b_i^z\right)^{\frac{1}{z}}\le \sum b_i$$ or $$\sum \left(\frac{b_i}{\sum b_k}\right)^z \le 1$$

But now $c_i = \frac{b_i}{\sum b_k}$ are positive with sum $1$, and $z>1$, so $$\sum c_i^z \le \sum c_i = 1$$

Thomas Andrews
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orangeskid
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  • https://math.stackexchange.com/questions/4094/how-do-you-show-monotonicity-of-the-ellp-norms – Joshua P. Swanson May 21 '21 at 22:29
  • @Joshua P. Swanson: Should be brought to the attention of the poster, so he would delete it promptly. – orangeskid May 21 '21 at 22:34
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    Meh, monotonicity of $\ell_p$ norms on sequence space is extremely well-known, but that's stated too generally for OP. This very concrete special case, which has a mildly easier proof I guess, is likely to be what someone else someday is actually searching for. I think it's fine as-is. – Joshua P. Swanson May 21 '21 at 22:37