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I had a problem of finding minima of a function $$f(x)=2^{x^2}-1+\frac{2}{2^{x^2}+1}$$

I solved it using AM-GM inequality, $$2^{x^2}-1+\frac{2}{2^{x^2}+1}$$ $$2^{x^2}+1+\frac{2}{2^{x^2}+1}-2$$ $$2^{x^2}+1+\frac{2}{2^{x^2}+1}\ge\ 2\sqrt2$$ $$2^{x^2}-1+\frac{2}{2^{x^2}+1}\ge\ 2\sqrt2-2$$

But in the solution answer was given as 1 and it was solved using differentiation, $$f'(x)=\frac{2x.ln2.2^{x^2}(2^{x^2}+1-\sqrt2)(2^{x^2}+1+\sqrt2)}{(2^{x^2}+1)^2}$$ $$2^{x^2}\ge1$$ $$2^{x^2}+1-\sqrt2\ge2-\sqrt2>0$$ At $x=0$ ,$f(x)$ is least.

Least value = $f(0)$ $=1$

I cannot understand how can there be two values by two different methods,please help me in the problem.

pranjal verma
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    AM-GM inequality is not always guaranteed to give you the greatest lower bound of a set, as such the two answers are not contradictory since AM-GM gives you something around $.8$. If the function is always at least $1$ then it is certainly greater than $.8$ – Ninad Munshi Oct 17 '19 at 11:14
  • @NinadMunshi used it as I have seen its use in solving problems like this. – pranjal verma Oct 17 '19 at 11:16

2 Answers2

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The fact is that the equality by AM-GM

$$2^{x^2}+1+\frac{2}{2^{x^2}+1}=\ 2\sqrt2$$

holds if and only if $\exists x$ such that

$$2^{x^2}+1=\frac{2}{2^{x^2}+1}$$

but

  • $2^{x^2}+1\ge 2$
  • $\frac{2}{2^{x^2}+1}\le 1$

therefore the value $2\sqrt 2$ is never reached and the inequality becomes

$$2^{x^2}+1+\frac{2}{2^{x^2}+1}>\ 2\sqrt2$$

user
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1

A much simpler method is to define $$y(x)=2^{x^2}$$ and $$g(y) = y-1+\frac{2}{y+1}$$

Such that $f(x) = g(y(x))$.

Then, rewrite g : $$g(y) = \frac{y^2+1}{y+1}$$

So, it's easy to see that g is increasing on $[1,+\infty)$, and $y\geq1$, therefore, the minimum of $f$ is $g(1)=1$ with $x=0$

Damien
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