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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.

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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? –  Elias 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

2 Answers 2

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.

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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
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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.

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