Seeking the maximal parameter value s.t. two-variable inequality still holds

Question

Consider the expression $$\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$ in two variables $\,a,b\,$ residing in $\,\mathbb R^{>0}$.

The arithmetic mean $\,\frac{a+b}2\,$ is a lower bound for it
$\big[$ one has $\,a^2b\le(2a^3+b^3)/3\,$ by AM-GM, do the same with $\,ab^2$, then add and divide by $2ab\,$$\big]$,
but $\,\max\{a,b\}\,$ is not
$\big[$ choose $\,(a,b)=(3,2)\,$ for instance, the expression then evaluates to $\,2\tfrac{11}{12}\,\big]$.

Both arithmetic mean ($x=1$) and the maximum ($x=\infty$) are instances of the Hölder mean $$\left(\frac{a^x+b^x}2\right)^{\frac 1x}\quad\text{with }\; x\,\in\,\{-\infty\}\cup\mathbb R\cup\{\infty\}$$ aka Power mean, known to be strictly increasing with $\,x\,$ if $\,a\ne b$, and my question is:

What is the maximal value of $\,x\,$ such that $$\left(\frac{a^x+b^x}2\right)^{\frac 1x}\;\leq\;\frac 12\left(\frac{a^2}b+\frac{b^2}a\right)$$ holds for all $\,a,b>0\,$?

Let's pick up the specific "max counter-example" $\,(a,b)=(3,2)\,$. The following plot screen-shot displays the zero, by courtesy of WolframAlpha:

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Returning to the general case we can at least state that $\,x_{max}\geqslant 5\,$:
After taking the fifth power of the corresponding expression and clearing denominators (to arrive at the LHS as of below) I could find a Certificate of positivity $$\begin{eqnarray} \left(a^3 + b^3\right)^5 -16a^5b^5\left(a^5 +b^5\right)\; & =\;\left(a+b\right)\left(a-b\right)^2\big[a^{12} + a^{11}b +2a^{10}b^2 +4a^9b^3 +8a^8b^4 \\ & +8a^6b^6 +8a^4b^8 +4a^3b^9 + 2a^2b^{10} +ab^{11} +b^{12}\big] \\ & + 3a^3b^3\left(a^2+b^2\right)\left(a+b\right)^3\left(a-b\right)^4 \end{eqnarray}$$ It shows also that the equality case holds iff $\,a=b$.

Answer 1

Not an answer, just some illustration. First, introducing:

$$y=\frac{a}{b}$$

We are looking at an inequality:

$$\left(\frac{y^x+1}2\right)^{\frac 1x}\;\leq\;\frac 12\left(y^2+\frac{1}y\right)$$

Plotting the difference with Wolfram Alpha for $x=1..10$ and for $y \in (0,20)$ we can see that the function seemingly has a maximum around $y=1$.

However, taking a closer look, it seems that for $x>9$ the function has two maxima around $1$ where the inequality breaks.

Thus you should probably search for $x$ between $9$ and $10$.

How to find the exact value and to prove the above range remains to be seen (of course, we can directly prove the number of maxima for $x=8$ and $x=9$ to see for ourselves if we want.

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An interesting picture if we take even smaller ranger for $y$:

Apparently $x=9$ is (or is close to) a borderline case.

Answer 2

Let $a=b-1$, so $\frac{1}{2}(\frac{a^2}{b}+\frac{b^2}{a}) =b+\frac{1}{2 (b-1)}+\frac{1}{2 b}-\frac{1}{2} < b-\frac{1}{3} $ true for $b>6.54138$.

Now $ \lim \limits_{x \to \infty} (\frac{a^x+b^x}{2})^{\frac{1}{x}} =(\frac{(b-1)^x+b^x}{2})^{\frac{1}{x}} = ((b-1)^x +b^x)^{\frac{1}{x}} *0.5^{1/x} = (b^x(1+(1-1/b)^x))^{\frac{1}{x}}*0.5^{\frac{1}{x}} = (b^x)^{\frac{1}{x}} (1+(1-1/b)^x)^{\frac{1}{x}} *0.5^x = b *(1+(1-\frac{1}{b})^x)^{\frac{1}{x}} *0.5^\frac{1}{x} = b$. assuming $b>2$.

So from $x_0$ every $x \geq x_0$ will have that $(\frac{(b-1)^x+b^x}{2})^{\frac{1}{x}} > b-\frac{1}{3}$.

And its not just $a=b-1$ , any $a$ you choose between $0.9b <a <b$, will have such $x_0$ that $x\geq x_0$ your inequality does not hold.

For instance $b=500$ so $\frac{1}{2}(\frac{a^2}{b}+\frac{b^2}{a}) = 499.502002 <499 \frac{2}{3} = 500-\frac{1}{3}$

Yet for all $x \geq 723 $ the term $(\frac{(b-1)^x+b^x}{2})^{\frac{1}{x}} \geq 499 \frac{2}{3}$.

The inequality is false when $a <b$ and $a$ "near" $b$ , and $x$ is sufficiently large.

Answer 3

The following inequality is true.

Let $a$ and $b$ be positive numbers.

Prove that $$\frac{1}{2}\left(\frac{a^2}{b}+\frac{b^2}{a}\right)\geq\sqrt[9]{\frac{a^9+b^9}{2}}.$$

Indeed, we need to prove that $$(a^3+b^3)^9\geq256(a^9+b^9)a^9b^9$$ or $$(a^3+b^3)^8\geq256(a^6-a^3b^3+b^6)a^9b^9.$$ Now, let $a^6+b^6=2ua^3b^3.$

Hence, $u\geq1$ and we need to prove that $$(2u+2)^4\geq256(2u-1),$$ which is AM-GM: $$(2u+2)^4=(2u-1+3)^4\geq\left(4\sqrt[4]{(2u-1)\cdot1^3}\right)^4=256(2u-1).$$ Done!

Answer 4

Here is a proof that the maximal $x$ must be $\le 9$.

We consider $a = 1 + u$, $b = 1$ and compute the Taylor development of $$ f(u) = \frac 12 \left( (u+1)^2 + \frac{1}{u+1}\right) - \left( \frac{(1+u)^x + 1}{2}\right)^\frac 1x $$ at $u = 0$.

From the geometric series we get $$ \frac 12 \left( (u+1)^2 + \frac{1}{u+1}\right) = 1 + \frac 12 u + u^2 + O(u^3) $$ and from the binomial series we get $$ (1+u)^x + 1 = 2 + xu + \frac{x(x-1)}{2} u^2 + O(u^3) $$ and then $$ \left( \frac{(1+u)^x + 1}{2}\right)^\frac 1x = 1 + \frac 12 u + \frac{x-1}{8}u^2 + O(u^3) $$ It follows that $$ f(u) = \frac{9-x}{8} u^2 + O(u^3) \text{ for } u \to 0 \, . $$

If $x > 9$ then $f(0) =f'(0) = 0$ and $f''(0) < 0$, i.e. $f$ has a strict local maximum at $u = 0$, so that the desired inequality does not hold for $a = 1+u$, $b=1$ with $u \ne 0$ sufficiently small.