Let us consider the following function :


In a particular Olympiad problem I have to prove that for any $a>0$,


So, I started up by assuming that the range of the function should be real numbers, because in case of complex numbers it would not make sense to speak of inequalities.

Then, the next set up was to conclude that $x$ cannot be less than $-1$ as it would not give real values for the fist term in $f(x)$.

And similarly, in the last term the sign of the denominator and the numerator must be the same which restricts $x$ from assuming values in $(-8,0)$. And hence $x$ should only have non negative values.

Now, I viewed $f(x)$ as a dot-product of vectors $$\left( 1, 1, \sqrt{\frac{ax}{ax+8}} \right) \quad \text{and} \quad \left( \frac{1}{\sqrt{1+x}}, \frac{1}{\sqrt{1+a}}, 1 \right)$$

Now the square of the dot product of these vectors must be less than the product of the square of the magnitudes of these vectors, which I guess is one way of Cauchy-Schwarz inequality.


$$f(x) < \left( \frac{1}{1+x}+\frac{1}{1+a}+1 \right) \left( 1+1+\frac{ax}{ax+8} \right)$$

At this stage, I can see that the as both the terms in the RHS of the inequality, decreases as $x$ and $a$ increases.

So the max value can be said to be $6$. But I could not tighten the bound to $4$.

I tried further by taking a term like

$\dfrac{1}{1+x}$ and making an inequality:

$\dfrac{1}{1+x} < \dfrac{1}{2\sqrt{x}}$ by applying AM-GM inequality

and then seeing $\dfrac{1}{\sqrt{x}}$ as a product and applying AM-GM to get $\dfrac{1}{\sqrt{x}} < \dfrac{1}{2} \left( 1 + \dfrac{1}{x} \right)$

But this couldn't tighten the upper bound.

Is there any way to proceed further by using Cauchy-Schwarz inequality without using any tools from calculus?

Raw picture

Basically , what I want to know is whether I can prove that f(x) will always be less than 2 by using cauchy-schwarz inequality and AM-GM inequality in the way I have tried to ?

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    $\begingroup$ The readability of this post is very poor, please try to improve it by using MathJax properly. $\endgroup$ – Jack D'Aurizio Apr 27 '17 at 18:17
  • $\begingroup$ You can type $\frac12$ using \frac{1}{2}. $\endgroup$ – user251257 Apr 27 '17 at 18:20
  • $\begingroup$ @JackD'Aurizio : I tried editing it , but couldn't find much, within my abilities to improve . Thought of replacing all fractions by frac , but that wouldn't have made much difference , I guess . $\endgroup$ – Mikhail Tal Apr 27 '17 at 19:23
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    $\begingroup$ Note that $$\frac{1}{\sqrt{\frac{ax}{8}+ax}}=\frac{\sqrt{8}}{3\sqrt{ax}}$$ which blows up at $x=0$. $\endgroup$ – Ng Chung Tak Apr 27 '17 at 19:58
  • $\begingroup$ @NgChungTak : There was a slight mistake in placing the indices .I have edited it , now it isn't blowing up I guess $\endgroup$ – Mikhail Tal Apr 28 '17 at 3:09

Let $x=\frac{2qr}{p^2}$ and $a=\frac{2pr}{q^2}$, where $p$, $q$ and $r$ are positives.

Hence, $\frac{8}{ax}=\frac{2pq}{r^2}$ and we need to prove that $$1<\frac{p}{\sqrt{p^2+2qr}}+\frac{q}{\sqrt{q^2+2pr}}+\frac{r}{\sqrt{r^2+2pq}}<2.$$ The left inequality.

By Holder: $$\left(\sum_{cyc}\frac{p}{\sqrt{p^2+2qr}}\right)^2\sum_{cyc}p(p^2+2qr)\geq(p+q+r)^3.$$ Thus, it remains to prove that $$(p+q+r)^3\geq\sum_{cyc}(p^3+2pqr),$$ which is obviously true because $(p+q+r)^3=\sum\limits_{cyc}(p^3+3p^2q+3p^2r+2pqr)$.

Right inequality.

We need to prove that $$\frac{1}{\sqrt{1+2x}}+ \frac{1}{\sqrt{1+2y}}+\frac{1}{\sqrt{1+2z}}<2$$ for positives $x$, $y$ and $z$ such that $xyz=1$.

Indeed, let $x=e^a$, $y=e^b$ and $z=e^c$.

Hence, $a+b+c=0$ and we need to prove that $\sum\limits_{cyc}g(a)<2$, where $$g(a)=\frac{1}{\sqrt{1+2e^{a}}}$$. We have $$g''(a)=\frac{e^a(e^a-1)}{\sqrt{1+2e^a)^5}},$$ which says that $g$ is a concave function for all $a\leq0$.

Thus, by the Vasc's LCF Theorem, it's enough to prove $\sum\limits_{cyc}g(a)<2$ for $b=a$.

Id est, we need to prove that $$\frac{2}{\sqrt{1+2x}}+\frac{1}{\sqrt{1+\frac{2}{x^2}}}<2$$ or $$2\sqrt{(x^2+2)(2x+1)}>x\sqrt{2x+1}+2\sqrt{x^2+2}$$ or $$6x^2-x+16>4\sqrt{(2x+1)(x^2+2)}$$ or $$36x^4-44x^3+177x^2-96x+224>0,$$ which is obvious.


  • $\begingroup$ The right inequality ? $\endgroup$ – Mikhail Tal Apr 29 '17 at 5:23
  • $\begingroup$ @Mikhail Tal See now. $\endgroup$ – Michael Rozenberg Apr 29 '17 at 5:44
  • $\begingroup$ Rozenberg : Nice observation and substitution ,xyz = 1 ...what is this LFC theorem that you are using ? $\endgroup$ – Mikhail Tal Apr 29 '17 at 7:37
  • $\begingroup$ @Rozenberg : Certainly your ways don't use anything from calculus apparently and I don't know if the techniques that you have used like holder's inequality and LFC theorem can somehow be intersecting with Cauchy -schwarz . But do you think there is any way in which we can use cauchy-schwarz to proceed towards the result ? $\endgroup$ – Mikhail Tal Apr 29 '17 at 7:39
  • $\begingroup$ @Rozenberg : Could you also tell me if there is any flaw in my method , which could only bound it upto 6 ? $\endgroup$ – Mikhail Tal Apr 29 '17 at 7:40

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