Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In my studies of various geometric inequalities I reached an inequality which seems true (numerically) but I cannot prove it. Let $p$, $q$, and $r$ be real numbers from the interval $(0,1)$. Let's also define the following function $$f({p})=\frac{\sqrt{1-p}}{(2-p)^2}$$ Prove (or disprove) that: $$ \frac{f(p)+f(q)+f(r)}{\sqrt{p q r}}\leq \frac{f(p)}{p\sqrt{p}}+\frac{f(q)}{q\sqrt{q}}+\frac{f(r)}{r\sqrt{r}} $$

I've tried Lagrange multipliers but the resulting equations do not seem tractable.

EDIT: The original question had the condition $p+q+r=2$ which apparently is not necessary, so I dropped it. I can prove that the inequality holds for $p=q$. A possible strategy is to try to establish monotonicity in one of the parameters under certain conditions. Unfortunately I can't manage the calculations.

share|cite|improve this question
I've noticed that this inequality seems to be true for other functions $f(x)$. Which suggests an additional question - what conditions are needed for $f(x)$ so that the inequality holds given the initial conditions. – ivan Dec 14 '12 at 7:28
Hi ivan, the inequality would not be contrary? – MathOverview Dec 15 '12 at 15:04
No, it is like this. – ivan Dec 15 '12 at 16:57
@ivan : of course, it is no coincidence that you treated the $p=q$ case. For a fixed $r$, and when we let $p$ and $q$ vary, numerically it seems that the minimum of the difference is attained when $p=q$. This is a familiar pattern in symmetrical inequalities : optimality is reached when the variables are equal. – Ewan Delanoy Dec 17 '12 at 8:51
I would be tempted to study $$g(p,q,r)= \frac{f(p)}{p\sqrt{p}}+\frac{f(q)}{q\sqrt{q}}+\frac{f(r)}{r\sqrt{r}}-\frac{f(p)+‌​f(q)+f(r)}{\sqrt{p q r}}\,.$$ It is clear that $g(1,1,1)=0$ (and even $g(p,1,1)\geq0$). If one could prove that $\frac{\partial}{\partial p} g(p,q,r)\leq 0$ on $(0,1]^3$, it would be sufficient (using the symmetry in $p$, $q$, $r$) to conclude on $(0,1]^3$, and then I guess the case with one or several of the other variables equal to $0$ could be handled separately. But the partial derivative does not seem to be very nice, as a maple computation indicates. – Sebastien B Dec 17 '12 at 13:39

This is a comment too long to fit in the usual format. Put $g(x)=\frac{f(x)}{x\sqrt{x}}$. Then the inequality to be shown is

$$ \frac{f(p)+f(q)+f(r)}{\sqrt{p q r}}\leq g( p ) +g( q ) +g( r ) \tag{1} $$

I can show this inequality in a special case, when $r=\frac{1}{10}$. Indeed, a stronger inequality holds in this case :

$$ \frac{f(p)+f(q)+f(r)}{\sqrt{p q r}}\leq g( r ) \tag{2} $$

To show (2), it will suffice to show the following four inequalities :

$$ \begin{array}{lc} \frac{f(p)}{\sqrt{p q r}}\leq \frac{9}{10} & (3) \\ \frac{f(q)}{\sqrt{p q r}}\leq \frac{9}{10} & (4) \\ \frac{f(r)}{\sqrt{p q r}}\leq \frac{9}{10} & (5) \\ 6 \leq g(r) & (6) \\ \end{array} $$

Consider the term $$T_1=\bigg(\frac{9}{10} (2-p)^2\bigg)^2pqr - (1-p) $$ Using the fact that $r=\frac{1}{10}$ and $q=(19/10)-p$, $T_1$ can be rewritten $$ T_1=\frac{673289}{10^8}+\frac{62373961}{10^8}(1-q)+\frac{29403}{80000}(1-q)^2+ (1-q)^3\Bigg(\frac{81}{1000}(1-p)^3 + \frac{1701}{5000}(1-p)^2 + \frac{48033}{100000}(1-p) + \frac{58887}{500000}\Bigg) $$ So $T_1$ is nonnegative, which yields (3). Interchanging $p$ and $q$, we obtain (4). We have $$ f ( r )=\frac{1}{(2-\frac{1}{10})^2} \sqrt{1-\frac{1}{10}}=\frac{300}{361\sqrt{10}} \tag{7} $$ and hence $$ \frac{f ( r )}{\sqrt{pqr}} = \frac{300}{361\sqrt{pq}} $$ The identity $$ pq-(\frac{10}{9} \times \frac{300}{361})^2=\frac{556001}{11728890}+(1-p)(1-q) $$ shows that $pq \geq (\frac{10}{9} \times \frac{300}{361})^2$, which yields (5). Finally, we deduce from (7) that $$ g ( r )=\frac{f ( r ) }{r\sqrt{r}}=\frac{300}{361\sqrt{10}} \times 10\sqrt{10}=\frac{3000}{361} $$ and this is indeed larger than $6$, which proves (6) and settles the $r=\frac{1}{10}$case.

share|cite|improve this answer
Nice. I tried to generalize this without using $p+q+r=2$ (which seems not to be necessary) but couldn't do it. – ivan Dec 17 '12 at 7:42
up vote 2 down vote accepted

I was able to prove this, finally. Here is a brief sketch of the proof. I will use the following simple fact:

Lemma. For positive numbers, if $a\geq b\geq c$ and $(x_1,x_2,x_3)\succ(y_1,y_2,y_3)$ then $ax_1+bx_2+cx_3\geq ay_1+by_2+cy_3\geq ay_i+by_j+cy_k$ where $(i,j,k)$ is an arbitrary permutation of $(1,2,3)$

Now notice that the function : $g(p)=f(p)/\sqrt{p}$ is decreasing in $(0,1)$. Assume $p\leq q\leq r$. Our inequality is equivalent to:$$\frac{g(p)}{p}+\frac{g(q)}{q}+\frac{g(r)}{r}\geq\frac{g(p)}{\sqrt{q r}}+\frac{g(q)}{\sqrt{p r}}+\frac{g(r)}{\sqrt{p q}}$$ Let's put $x_1=1/p, x_2=1/q$, $x_3=1/r$ and $y_1=(x_1+x_2)/2, y_2=(x_1+x_3)/2, y_3=(x_2+x_3)/2$. Notice that $x_1\geq x_2\geq x_3$, $y_1\geq y_2\geq y_3$ and $(x_1,x_2,x_3)\succ(y_1,y_2,y_3)$. Applying the lemma for $a=g(p), b=g(q)$ and $c=g(r)$ ($a\geq b\geq c$ because $g(x)$ is decreasing) we get: $$ ax_1+bx_2+cx_3\geq ay_3+by_2+cy_1=a\frac{x_2+x_3}{2}+b\frac{x_1+x_3}{2}+c\frac{x_1+x_2}{2}\geq a\sqrt{x_2 x_3}+b\sqrt{x_1 x_3} + c\sqrt{x_1 x_2} $$

and this is exactly what we are trying to prove.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.