Why does AM>GM when applied on functions gives the absolute minima.

Question

• In some cases we use the relation AM>GM to find the minima for example take $f(x)=x+\frac1x$ $[x\gt 0]$ using the result AM>GM we can find the minima as $2$.It is the same minima which we get if we use the methods of derivative.
• But why do we get the same minima why not say we get $f(x)>1$.This statement ($f(x)>1$) is not false but $1$ is just not the absolute minima.Can we prove that the relation AM>GM will always give the absolute minima in the given domain? (Note-I am not asking for the proof of result AM>GM.)
• Hence as we are getting the absolute minima using the result AM>GM I conclude that there is some kind of relation between applying the result $AM>GM$ and using the methods of derivative.Is there any intuitive way to understand that relation?

Answer 1

Using AM-GM not always gives an extreme value. We need also to save the case of the equality occurring. In AM-GM it happens for equality case of all variables, which not always good.

Also, under AM-GM we have convexity of $\ln$, which has a relation with a second derivative.

Also, there are very many methods to find an extreme value with derivatives and these methods have no any relation with AM-GM.

For example.

Let we need to find a maximal value of $(a^2-ab+b^2)(a^2-ac+c^2)(b^2-bc+c^2)$ by AM-GM, where $a$, $b$ and $c$ are non-negatives such that $a+b+c=3$.

We see that $(a,b,c)=(2,1,0)$ gives a value $12$.

We'll prove that $12$ is a maximal value.

We can not use AM-GM here in the following form. $$\prod_{cyc}(a^2-ab+b^2)\leq\left(\frac{\sum\limits_{cyc}(a^2-ab+b^2)}{3}\right)^3$$ because it does not save the case of equality occurring.

For $(a,b,c)=(2,1,0)$ we obtain: $$a^2-ab+b^2=3,$$$$a^2-ac+c^2=4$$ and $$b^2-bc+c^2=1,$$ which is not good because we need $$a^2-ab+b^2=a^2-ac+b^2=b^2-bc+c^2,$$ which is wrong and it says that we got a wrong inequality: $$\left(\frac{\sum\limits_{cyc}(a^2-ab+b^2)}{3}\right)^3\leq12,$$ which easy to understand after checking $(a,b,c)=(2,1,0).$

By the way, the following solution saves the case of equality occurring.

Let $a\geq b\geq c$.

Thus, $$(a^2-ab+b^2)(a^2-ac+c^2)(b^2-bc-c^2)\leq(a^2-ab+b^2)a^2b^2=$$ $$=((a+b)^2-3ab)a^2b^2\leq(9-3ab)a^2b^2=12(3-ab)\left(\frac{ab}{2}\right)^2\leq$$ $$\leq12\left(\frac{3-ab+\frac{ab}{2}+\frac{ab}{2}}{3}\right)^3=12.$$ Id est, we solved this problem by AM-GM.

Actually, using AM-GM was possible because for $(a,b,c)=(2,1,0)$ we have $$3-ab=\frac{ab}{2}=\frac{ab}{2}.$$ We can use the Lagrange Multipliers method here (id est, we can use derivatives),

but it not so easy here.