Finding the minima of a function in two ways

Question

I found a simple question:- Find the minimum value of $x^2+\frac{1}{x^2}+4$ where $x\in R$

Now,By AM-GM inequality $$x^2+\frac{1}{x^2}\ge2$$$$x^2+\frac{1}{x^2}+4\ge6$$ Thus min. value $=6$

Second solution:-$$({x+\frac{1}{x}})^2\ge0$$$$\Rightarrow x^2+\frac{1}{x^2}+2\ge 0$$ $$\Rightarrow x^2+\frac{1}{x^2}+4\ge2$$ So min. value$=2$

Now answer from the book says min.$=2$ But I find a contradiction in the Second solution.

If min$=2$ $$x^2+\frac{1}{x^2}+2=0$$$$\Rightarrow x^2+\frac{1}{x^2}=-2$$ Sum of two squares can be negative only if $x^2$ is negative or $x$ is imaginary which is a contradiction to given statement $x\in R$. So answer must be 6.

Am I making a mistake somewhere or am I correct??

Answer 1

By AM-GM $$x^2+\frac{1}{x^2}+4\geq2\sqrt{x^2\cdot\frac{1}{x^2}}+4=6.$$ The equality occurs for $x=1$, which says that $6$ is a minimal value.

Now, about your question.

Indeed, we need to find a minimum of $$\left(x+\frac{1}{x}\right)^2+2,$$ but we can not say, that the minimum is $2$, because $x+\frac{1}{x}\neq0$ for all real $x$.

Id est, the equality in the inequality $$x^2+\frac{1}{x^2}+4\geq2$$ does not occur, but the inequality $$x^2+\frac{1}{x^2}+4\geq2$$ is true.

All this says that the minimum greater than $2$ and indeed, the minimum is $6$.

Answer 2

with $$f(x)=x^2+x^{-2}+4$$ we get $$f'(x)=2x-2x^{-3}$$ $$f''(x)=2+6x^{-4}$$ solving $$f'(x)=0$$ we obtain $x=-1$ or $x=1$ with the second derivative test we get that is clearly a minimum. thus $$f(1)=f(-1)=6$$

Answer 3

By replacing $t=x^2$ we notice that the min of $x^2+\frac{1}{x^2}$ over $R$ is the same as min of $t+\frac{1}{t}$ over positive $t$. The latter min is 2 attained at $t=1$.

Proving that $f(x) \geq a$ does not necessarily mean that $a$ is a minimum. It only means that the min is no less than $a$. Either it is $a$ or some bigger number. That is why there is no contradiction in your estimates. You showed that $f(x)\geq 6$ with some manipulation and used other tools to also show $f(x) \geq 2.$ these together say that the min is $6$ or larger -- and not 2 on earth! Since $6$ is attained at $x=1$ this becomes the min.