Prove that $1 < \frac{1}{\sqrt {1+x}} +\frac{1}{\sqrt {1+a}} + \sqrt{\frac{ax}{ax+8}} < 2$ for $a, x > 0$

Question

Let $a>0$, show that for $x>0$, $1<f(x)<2$, where $$f(x)=\frac{1}{\sqrt{1+x}}+\frac{1}{\sqrt{1+a}}+\sqrt{\frac{ax}{ax+8}}.$$

Source (added 20221017): It was the last question in math exam of Jiangxi Province, 2008 China's national college entrance exam (a.k.a. Gaokao). It was said that no examinee proved $f(x) < 2$. It was seen as one of the hardest math problems in GaoKao history. It attracted much attention in the community of high school math or Olympiad math. Perhaps there are more than 10 proofs. By the way, it was reported that in 2021 more than 10 million students took Gaokao.

I could take the derivative, find the maximum of the function and conclude. But that involves heady algebra and you have to solve a high degree polynomial. Is there an easier way simply using inequalities? Thanks, much appreciated

Answer 1

note the 3 items all less then 1, with $\sqrt{u} \ge u \cap 0<u<1$ so we have:

$f(x)> \dfrac{1}{1+x}+\dfrac{1}{1+a}+\dfrac{ax}{ax+b}=g(x) $ here we use $b$ to take 8.

with $\sqrt{u}+\sqrt{v} \ge \sqrt{u+v}$, we have:

$f(x)<\dfrac{1}{1+\sqrt{x}}+\dfrac{1}{1+\sqrt{a}}+\dfrac{\sqrt{ax}}{\sqrt{ax}+\sqrt{b}}=\dfrac{1}{1+x'}+\dfrac{1}{1+a'}+\dfrac{a'x'}{a'x'+b'} ,x'=\sqrt{x},a'=\sqrt{a},b'=\sqrt{b}$

so we only need to find the bound of $g(x)$ which is much easier.

$g'(x)=0 \implies x_1=\sqrt{\dfrac{b}{a}}$ and there is only one critical point.

so you only need to prove $1 \le g(x_1) \le 2$ and the with $ g(0)=g(+\infty)=1+\dfrac{1}{1+a}$, the proof is done.

Answer 2

Here is my proof for $f(x) < 2$ without calculus:

It suffices to prove that, for all $a, b, c > 0$ with $abc = 8$, $$\frac{1}{\sqrt{1 + a}} + \frac{1}{\sqrt{1 + b}} + \frac{1}{\sqrt{1 + c}} < 2.$$

WLOG, assume that $a \le b \le c$.

We split into two cases:

1. $a + b \ge 6$:

We have $b \ge 3$. Thus, $$\mathrm{LHS} < 1 + \frac{1}{\sqrt{1 + 3}} + \frac{1}{\sqrt{1 + 3}} = 2.$$

2. $a + b < 6$:

Using AM-GM, we have $$\frac{1}{\sqrt{1 + a}} = \frac{\sqrt{1 + a}}{1 + a} \le \frac{\frac{(1 + a) + 1}{2}}{1 + a} = \frac{a + 2}{2 + 2a}$$ and $$\frac{1}{\sqrt{1 + b}} = \frac{\sqrt{1 + b}}{1 + b} \le \frac{\frac{(1 + b) + 1}{2}}{1 + b} = \frac{b + 2}{2 + 2b}.$$

It suffices to prove that $$\frac{a + 2}{2 + 2a} + \frac{b + 2}{2 + 2b} + \frac{1}{\sqrt{1 + c}} < 2$$ or $$\frac{a}{2 + 2a} + \frac{b}{2 + 2b} > \frac{1}{\sqrt{1 + c}}.$$

Using AM-GM, it suffices to prove that $$ 2\sqrt{\frac{a}{2 + 2a} \cdot \frac{b}{2 + 2b}} > \frac{1}{\sqrt{1 + c}}$$ or $$4\cdot \frac{a}{2 + 2a} \cdot \frac{b}{2 + 2b} > \frac{1}{1 + c}$$ or $$\frac{ab(7 - a - b)}{(1 + a)(1 + b)(ab + 8)} > 0$$ which is true.

We are done.

Answer 3

HINT: show that $$\lim_{x \to \infty}f(x)=\frac{\sqrt{1+a}+1}{\sqrt{1+a}}$$ compute the solutions of the equation $$8a\sqrt{x+1}(x+1)-\sqrt{ax(ax+8)}(ax+8)=0$$