# Trying to grasp the concept of homogeneous inequalities

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?

• 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. – Macavity Jul 26 '19 at 5:50

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!