Finding the minimum value of function ${27𝑎^6}$ +$\frac{1}{(𝑎^2−𝑏^2−𝑐^2+2𝑏𝑐)}$

Question

When I put this function in GeoGebra calculator suite, I am getting the minimum value 4 but when I tried to solve, the minimum value I'm getting is different:

My attempt:

${27𝑎^6}$ +$\frac{1}{(𝑎^2−𝑏^2−𝑐^2+2𝑏𝑐)}$ can be written as ${27a^6}$+$\frac{1}{a^2-(b-c)^2}$

i.e. ${27a^6}$+$\frac{1}{(a+b-c)(a-b+c)}$

Observing, we know that, the minimum occurs at b=c.

so, the function gets reduced to ${27a^6}$+$\frac{1}{a^2}$

Using AM-GM inequality, we get,

$\frac{{27a^8}+1}{2{a^2}}$ ≥ √(${27a^4}$)

or, ${27a^8}$+1 ≥ 6√3 ${a^4}$

or, ${27a^8}$-6√3 ${a^4}$+1 ≥ 0

or,${(3√3 {a^4}-1)^2}$ ≥ 0

or, ${a^4}$ ≥ $\frac{1}{3√3}$

∴ ${a^2}$ ≥ $\frac{1}{√(3√3)}$

Now, substituting the value of ${a^2}$ to the original equation, we get,

${27a^6}$+$\frac{1}{a^2}$ = ${{(3a^2)}^3}$ + $\frac{1}{a^2}$

≥($\frac{3}{√(3√3)})^3$ + √(3√3)

≥ 2*${3^\frac{3}{4}}$

≥4.559014113909555283987126503927272012380640693323861911667731922

∴${27a^6}$+$\frac{1}{a^2}$ ≥4.6

This implies that the minimum value of the function is 4.6 while it is 4. How? Please point out my error and also suggest your ways to solve this. GEOGEBRA GRAPH

Answer 1

You have to minimize the function $$f(a) = 27a^6 + \frac{1}{a^2}$$ with respect to $a$. For that, you can simply take the first derivative and equate it to $0$.

Following these steps, you first get that $$\frac{\partial f(a)}{\partial a} = 162a^5-\frac{2}{a^3}.$$ From there, it follows that $$ a^* = \left(\frac{1}{81}\right)^\frac{1}{8}=\left(\frac{1}{3}\right)^\frac{1}{2}.$$

Then, substituting $a^* $ in $f(a)$, you get $$f(a^*) = \frac{27}{3^3} + 3 = 4.$$

To make sure that it corresponds to the minimum, you can take the second derivative and check that it is positive. If you do so, you get that $$\frac{\partial^2 f(a)}{\partial a^2} = 810 a^4+\frac{5}{a^4},$$ which is indeed positive for any value of $a$, thus concluding the proof.

Answer 2

AM-GM generally works best when you get a constant on the other side

$$ {1\over 4}(27a^6+{1\over 3a^2}+{1\over 3a^2}+{1\over 3a^2})\ge\sqrt[4]{1} $$

So $$27a^6+{1\over a^2}\ge4$$