Proof or references for strengthened AM-GM

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

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

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.

• 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. Commented Jan 1, 2023 at 23:55
• See this maybe: math.stackexchange.com/questions/2603759/… Commented Jan 1, 2023 at 23:59
• @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 $b$. Commented Jan 2, 2023 at 0:30
• No problem, found by googling "Bounds on the difference between AM and GM" Commented Jan 2, 2023 at 0:49
• 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 $y=0$. Commented Jan 2, 2023 at 5:05

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.

• Welcome to MSE. It is in your best interest that you type your posts (using MathJax) instead of posting links to pictures. Commented Jan 2, 2023 at 13:02
• Rehman, would you mind if I edited your answer in explicitly into your answer rather than as an image and then accept it? Commented Jan 2, 2023 at 13:37
• yes I accept. Dont warry, thats not problem for me Commented Jan 2, 2023 at 14:05