Proving an inequality involving real powers

Question

Q: For $p\in\mathbb{R}$ such that $2<p<\infty$, and $a,b>0$, prove the following

$$a^p+b^p<(a^2+b^2)^{p/2}$$

My attempt: I say it's equivalent to proving that $(a^p+b^p)^{1/p}<(a^2+b^2)^{1/2}$. Then I did binomial expansions on both sides of the inequality. Both the expansions have $a+b$ so I canceled them. For the rest of the terms, I thought I'd get geometric series on both sides and show that the series on the left is smaller than that of the right by using related sum formulas. But what I found were not quite geometric. For the left side

$$\frac{1}{p}a^{1-p}b^p+\frac{1-p}{2p^2}a^{1-2p}b^{2p}+\frac{(1-p)(1-2p)}{6p^3}a^{1-3p}b^{3p}+\ldots$$

While for the right

$$\frac{1}{2}a^{-1}b^2-\frac{1}{8}a^{-3}b^4+\frac{1}{16}a^{-5}b^6+\ldots$$

How to proceed to prove the inequality?

Answer 1

Thanks to @Michael Rozenberg for typing out this idea of mine: $$\frac{a^p+b^p}{(a^2+b^2)^{\frac{p}{2}}}=\left(\frac{a^2}{a^2+b^2}\right)^{\frac{p}{2}}+\left(\frac{b^2}{a^2+b^2}\right)^{\frac{p}{2}}\le\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=1,$$

where I have used the fact that $x^q\le x$ when $x\in]0,1]$ and $q\geq1$.

Answer 2

Let $a^2=x$, $b^2=y$ and $\frac{p}{2}=k.$

Thus, we need to prove that $$(x+y)^k\geq x^k+y^k,$$ which is Karamata for a convex function $f(t)=t^k$.

Indeed, let $x\geq y$.

Thus, $$(x+y,0)\succ(x,y),$$ which gives $$(x+y)^k+0^k\geq x^k+y^k,$$ which is our inequality.

Answer 3

First you can reduce the inequality to $x=\frac ba >0$

$$(1+x^p)^{\frac 1p} < (1+x^2)^{\frac 12} \Leftrightarrow \frac 1p\log(1+x^p)<\frac 12 \log(1+x^2)$$

by noting that $(a^p+b^p)^{\frac 1p}= a\left(1 +\frac{b^p}{a^p}\right)^{\frac 1p}$.

Now, consider for a fixed positive $x$ the function

$$f(p) = \frac 1p \log(1+x^p)$$

Differentiating wrt. $p$ gives $$\frac d{dp}(f(p))= \frac{1}{p^2(1+x^p)}\left(x^pp\log x -(1+x^p)\log(1+x^p)\right)$$ $$=\frac{1}{p^2(1+x^p)}\left(x^p\log\frac{x^p}{1+x^p}-\log(1+x^p)\right)<0$$

So, $f$ is strictly decreasing and, hence, the inequality holds true.

Note, that here is shown more:

$$\boxed{a^p+b^p<(a^q+b^q)^{p/q}}\text{ for } 0<q<p \text{ and }a,b>0$$