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

Let $$a>0$$, show that for $$x>0$$, $$1, where $$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

• Have you tried yet to take the derivative anyway? Who knows? Maybe something unexpected will happen, like, maybe the function is monotonic... Feb 9, 2016 at 0:15
• the function isn't monotonic Feb 9, 2016 at 0:20
• tried taking the derivative, lead to nowherer Feb 9, 2016 at 0:48

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.

• When you do one over, the inequality flips. So you have $f>g$ twice.
– Max
Jun 7, 2021 at 7:01
• Shouldn't the sign of inequality be reversed in the 4th line ? Jun 7, 2021 at 7:17
• Yes, $\sqrt{1 + x} < 1 + \sqrt x$ leads to $\frac{1}{\sqrt{1 + x}} > \frac{1}{1 + \sqrt x}$, so this answer for $f(x) < 2$ is incorrect. Oct 17, 2022 at 0:50

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, $$\mathrm{LHS} < 1 + \frac{1}{\sqrt{1 + 3}} + \frac{1}{\sqrt{1 + 3}} = 2.$$

1. $$a + b < 6$$:

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.

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

• yea, this is right, but how could this help, the function is not monotonic Feb 9, 2016 at 0:50
• yes i' m working Feb 9, 2016 at 0:51
• Thanks, for the tireless effort, I truly appreciate your work Feb 9, 2016 at 0:59
• why the $-1$? no one has a better solution Feb 9, 2016 at 11:27
• I just vote up to cancel that Feb 9, 2016 at 14:12