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let $a,b,c\ge 0$,and such $a+b+c=1$,show that $$\sqrt{1-ab}+\sqrt{1-bc}+\sqrt{1-ac}\ge2\sqrt{2}$$

maybe can use C-S to solve it.

My attempt is $$\sum_\mathrm{cyc}\sqrt{a^2+b^2+c^2+ab+2bc+2ac}\ge 2\sqrt{2}$$ then I want to square, but it's ugly.

clarification: I think that this question is not a duplicate of that question. The equality case for this question is $(1/3, 1/3, 1/3)$, while $(1, 1, 0)$ for that question. If we want solutions without using calculus (e.g. typically for contest), the approaches for them can be different. For example, my approach for that question does not work for this question because of the different equality case.

I noticed that in comment, Moonshine thought as well "It seems to be different, although the idea may be similar..."

River Li
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wightahtl
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2 Answers2

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Your expectation that it would use Cauchy ineq. is probably correct(maybe Jensen) but I won't because that kind of approach is much like ad-hoc.

Let $f(a,b,c)=\sqrt{1-ab}+\sqrt{1-bc}+\sqrt{1-ca}$.

Note that $f$ is continuous function defined on a compact set so this should have a maximal value.

Now I will prove a lemma.

Lemma 1. If $c\le 1/2$ and $(a,b,c)$ maximizes $f$, $a=b$.

proof. writing $b=1-c-a$, $0=2\cdot \frac{df}{da}=c\left(\frac{1}{\sqrt{1-ac}}-\frac{1}{\sqrt{1-bc}}\right)+\frac{a-b}{\sqrt{1-ab}}$ is required.

Assume $a>b$. We claim that $\frac{1}{\sqrt{1-ac}}-\frac{1}{\sqrt{1-bc}}<2(a-b)$.

Multiplying $\frac{1}{a-b}\left(\frac{1}{\sqrt{1-ac}}+\frac{1}{\sqrt{1-bc}}\right)$, this is equivalent to $\frac{c}{(1-ac)(1-bc)}<2\left(\frac{1}{\sqrt{1-ac}}+\frac{1}{\sqrt{1-bc}}\right)$ and since $ac,bc\le 1/4(\because a+b+c=1)$ and $c\le 1/2$, enough to show the following:

$\frac{1}{2}\cdot (\frac{4}{3})^2<2\cdot 2$

and this is obviously true. From this claim, we got contradiction.

Assuming $b>a$ similarly gives contradiction. Thus, $a=b$. □

Now, note that for any pair of $(a,b,c)$, at least two of the elements should be $\le 1/2$, thus we can apply Lemma 1 twice to get $a=b=c(=1/3)$.

Now, we know that $a=b=c=1/3$ is the only possible candidate of maximum. Put it in the original equation, then you get $2\sqrt{2}$.

Now we are done with the proof

... so why isn't this an ad-hoc?

This is a finite-variable optimization problem. (This kind of argument works for infinite dimension as well, but that case you need to be more careful about compactness, differentiability, etc.)

In optimization problem, what we want to do at first sight is to differentiate with respect to one variable, as we learned in calculus.

However, we cannot do that directly and the simplest thing to try is to fix all the variables but only two and see what happens, so I did that. You see $c\le 1/2$ in the lemma, but I just added that condition after calculating out the leftover; even if that did not hold, I wouldn't have worried much. Also, observing that equality holds when $a=b=c$ shows that this approach is very tempting.

Moonshine
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Here is a proof using isolated fudging.

Fact 1: Let $x, y\ge 0$ with $x + y \le 1$. Then $$\frac{\sqrt{1 - xy}}{2\sqrt 2} \ge - \frac{1}{12} x^2 - \frac{1}{12}y^2 - \frac13 xy + \frac{5}{48}x + \frac{5}{48}y + \frac{23}{72}.$$ (The proof is given at the end.)

Using Fact 1, we have \begin{align*} &\sum_{\mathrm{cyc}} \frac{\sqrt{1 - ab}}{2\sqrt 2}\\[6pt] \ge{}& \sum_{\mathrm{cyc}} \left(- \frac{1}{12} a^2 - \frac{1}{12}b^2 - \frac13 ab + \frac{5}{48}a + \frac{5}{48}b + \frac{23}{72}\right)\\[6pt] ={}& - \frac16 (a + b + c)^2 + \frac{5}{24}(a + b + c) + \frac{23}{24}\\[6pt] ={}& 1. \end{align*} The desired result follows.

We are done.

$\phantom{2}$

Proof of Fact 1:

We have \begin{align*} &\mathrm{LHS} - \mathrm{RHS}\\[6pt] ={}& -\frac43\left(\frac38 - \frac{\sqrt{1 - xy}}{2\sqrt 2}\right)^2 + \frac{1}{12}(x + y)^2- \frac{5}{48}(x + y) + \frac{5}{144}\\[6pt] \ge{}& -\frac43\left(\frac38 - \frac{\sqrt{1 - (x+y)^2/4}}{2\sqrt 2}\right)^2 + \frac{1}{12}(x + y)^2- \frac{5}{48}(x + y) + \frac{5}{144}\\[6pt] ={}& \frac{\sqrt{2(4 - p^2)}}{8} + \frac18 p^2 - \frac{5}{48}p - \frac{23}{72}\\[6pt] \ge{}& 0 \end{align*} where $p := x + y \in [0, 1]$, and we use $xy \le (x + y)^2/4$, and $$\left(\frac{\sqrt{2(4 - p^2)}}{8}\right)^2 - \left(\frac{5}{48}p + \frac{23}{72} - \frac18 p^2\right)^2 = \frac{12p + 119 -36p^2}{20736}(3p - 2)^2 \ge 0.$$

We are done.

River Li
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