Innate Inequality Part II

The minimum value is $\boxed{\frac{3\sqrt{3}}{2}}$ .

For an arbitrary positive real number $x<1$ , we have: $\frac{x}{1-x^{2}}\geq\frac{3\sqrt{3}}{2}x^{2}$ (1)

Indeed, (1) holds true because it is equivalent to: $x(\sqrt{3}x-1)^{2}(\sqrt{3}x+2)\geq0$

(The equality holds when $x=0$ or $x=\frac{1}{\sqrt{3}}$ ).

Apply this, we have: $\sum_{i=1}^{101}{\frac{a_i}{1-a_i^{2}}}\geq\sum_{i=1}^{101}{\frac{3\sqrt{3}}{2}a_i^{2}}$

Thus: $\sum_{i=1}^{101}{\frac{a_i}{1-a_i^{2}}}\geq\frac{3\sqrt{3}}{2}$ (2)

We can show a set of $(a_1, a_2, ..., a_{101})$ that satisfies the equality of (2), for example: $a_1=a_2=a_3=\frac{1}{\sqrt{3}}$ ; $a_4=a_5=...=a_{101}=0$ . This ends my proof.

The answer to this question is wrong!! I must confess, I spent way too long trying to figure out how to use Jensen's Inequality, or the AM-GM inequality or Cauchy-Schwarz to solve it. But nothing worked, and now I finally see why. Let's disprove the answer with a counterexample:

For $1 \le i \le 80$ , let $a_i = 0.11$ ; for $81 \le j \le 99$ let $a_j = 0.04$ ; let $a_{100} = 0.024$ and let $a_{101} = 0.032$ . Then plugging in the numbers gives $\sum_{i=1}^{101} a_i^2 = 1$ , and $\sum_{i=1}^{101} \frac{a_i}{1-a_i^2} = 9.725... < 10.15 = \frac{101\sqrt{101}}{100}$ Hence, the minimum value is not $\frac{101\sqrt{101}}{100}$ .

So why doesn't Jensen's Inequality work? Let $b_i = a_i^2$ . Then we are really trying to find the minimum of the following sum: $\sum_{i=1}^{101} \frac{\sqrt{b_i}}{1-b_i}$ , where $\sum_{i=1}^{101} b_i = 1$ . The problem is, Jensen's Inequality requires the function to be concave upward for the entire interval $(0,1)$ . However, the second derivative of $\frac{\sqrt{x}}{1-x}$ is: $\frac{3x^2+6x-1}{4x\sqrt{x}(1-x)^3}$ Therefore, it is not always concave upward, so we can't use Jensen's Inequality.

I honestly don't know all these complicated laws and equations like all the other geniuses on this site does, so I'll just post my way of solving this problem.

My method has its basis mainly on guesswork. My assumption was that when a1=a2=a3=......=a100=a101, the given equation will be minimal. In order to prove my assumption, I set a vastly simplified scenario of a1/(1-a1^2) + a2/(1-a2^2), and tried two examples where a1 = a2 = 1/sqrt(101), and where a1 = lim(x->0+)x and a2=lim(x->1-)x

The first instance gave me a result of (101 * sqrt101)/50, whereas the second gave a result of lim(x->inf)x

Therefore, I concluded that the result will be minimal when all the values of a were the same, hence obtaining the result of 101sqrt(101)/100 for the given equation.

In this case, a = 101, b = 101 and c = 100, therefore a - b + c = 101 - 101 + 100 = 100

HINT: Apply Jensen's Inequality by taking $f(x) = x/(1-x^2)$ .