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This other question includes the following strengthened version of the arithmetic-mean geometric-mean inequality.

\begin{equation} \label{1}\tag{1} \dfrac{a+b}{2} - \sqrt{ab} \geq \dfrac{1}{16 \max \left\lbrace a , b \right\rbrace} \left( a - b \right) ^{2} . \end{equation}

I'm wondering if anyone has a 1) proof of this and 2) a citation for it? My initial thought was to use Lagrange multipliers here, but the max on the bottom makes this approach a bit annoying.

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  • $\begingroup$ You can remove the max by making an explicit substitution that breaks the symmetry between $a$ and $b$, say $a = x, b = x + y$ and $x, y \ge 0$. I don't know if this is helpful though. $\endgroup$
    – Qiaochu Yuan
    Jan 1 at 23:55
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    $\begingroup$ See this maybe: math.stackexchange.com/questions/2603759/… $\endgroup$
    – TheBestMagician
    Jan 1 at 23:59
  • $\begingroup$ @TheBestMagician Ah, yes thank you. The Aldaz paper referenced there gets an inequality that is tighter than the one in question for all $a$ and $b$. $\endgroup$
    – JoshuaZ
    Jan 2 at 0:30
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    $\begingroup$ No problem, found by googling "Bounds on the difference between AM and GM" $\endgroup$
    – TheBestMagician
    Jan 2 at 0:49
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    $\begingroup$ As an aside, the approach suggested by Qianchu works. We want to show that (multiply by denominator and shift terms around) $16x^2 + 24xy + 7y^2 \geq 16(x+y)\sqrt{x(x+y)}$. This is true as $(16x^2+24xy+7y^2)^2 - (16(x+y))^2(x(x+y)) = 32x^2y^2 + 80xy^3 + 49y^4$. Equality holds (in non-negative) iff $y=0$. $\endgroup$
    – Calvin Lin
    Jan 2 at 5:05

1 Answer 1

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Here is an image version of the solution

Written out here:

We first break symmetry by assuming that $a \geq b$. We need to show then that $\frac{a+b-2\sqrt{ab}}{2} \geq \frac{\left(a-b\right)^2}{16a}.$

This is equivalent after clearing denominates and factoring the right hand side to $$8a(\sqrt{a}-\sqrt{b})^2 \geq (a-b)^2$$ which is equivalent to $$8a\frac{(\sqrt{a}-\sqrt{b})^2}{(a-b)^2} \geq 1$$ which is equivalent to $$8a\frac{1}{(\sqrt{a}+\sqrt{b})^2} \geq 1.$$

So we are done if we can show that $8a \geq (\sqrt{a}+\sqrt{b})^2.$

By the usual AM-GM inequality in the form $(x+y)^2 \leq2(x^2 +y^2)$, and using that $a \geq b$ we have $8a \geq 4a \geq 2(a+b) \geq (\sqrt{a}+\sqrt{b})^2$ and we are done. Note that this proof also shows that we could get a stronger constant in the original, replacing $16$ with $8$ in our original inequality.

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    $\begingroup$ Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. $\endgroup$
    – José Carlos Santos
    Jan 2 at 13:02
  • $\begingroup$ Rehman, would you mind if I edited your answer in explicitly into your answer rather than as an image and then accept it? $\endgroup$
    – JoshuaZ
    Jan 2 at 13:37
  • $\begingroup$ yes I accept. Dont warry, thats not problem for me $\endgroup$
    – Rehman
    Jan 2 at 14:05

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