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Given $a,b\in \mathbb{R}^+$, prove the following lower bound for the AM-GM difference:

$$\frac{a+b}{2} - \sqrt{ab} \geq \frac{(a-b)^2(a+3b)(b+3a)}{8(a+b)(a^2+6ab+b^2)} \tag 1$$ (Hint: setting $b=1,a=t^2,t>0$, you get a polynomial inequality that can be proved by way of factorization).

I understand that if you have an inequality $R>L$ with variables $x_1,\dots, x_n$ and you can find some $t\in \mathbb{R}$ such that for the change of variables $x_i'= t x_i$ you can find a common factor $t(R-L)>0$, then you may assume, for example, that for some $t_0$ you get $L=1$, so that the rest of the proof consists in showing $R>1$.

But how can I begin to prove $(1)$ knowing that it's homogeneous? Should I try to find a change of variables such that $\frac{a'+b'}{2} + \sqrt{a'b'}= \lambda$ for some $\lambda \in \mathbb{R}$? I'm very confused and I don't understand the "hint" at all (if $b=1$, aren't we losing solutions?).

Also, when proving inequalities we don't know if they're homogeneous, is it advisable to determine first whether they are?

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  • $\begingroup$ Homogeneous implies scaling by any positive real does not affect the inequality. Now if that is so, you can always choose a scale to make any one of the variables $=1$. This reduces the number of variables in the problem without affecting generality, hence is a better situation. Usually it is a good idea to check for homogeneity as you can set a variable’s value, or the sum or product of variables to some desired value etc. $\endgroup$ – Macavity Jul 26 '19 at 5:50
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It means the following:

Let $a=t^2b$, where $t>0$.

Thus, we need to prove that: $$\frac{t^2b+b}{2}-tb\geq\frac{(t^2b-b)^2(t^2b+3b)(b+3t^2b)}{8(t^2b+b)(t^4b^2+6t^2b^2+b^2)}$$ or $$b\left(\frac{t^2+1}{2}-t\right)\geq\frac{b^4(t^2-1)^2(t^2+3)(1+3t^2)}{8b^3(t^2+1)(t^4+6t^2+1)}$$ $$\frac{t^2+1}{2}-t\geq\frac{(t^2-1)^2(t^2+3)(1+3t^2)}{8(t^2+1)(t^4+6t^2+1)}$$ or $$(t-1)^6\geq0$$ and we are done!

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