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Let $x,y,z\ge 0$, and $x+y+z=2$, show that $$\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\ge 2.$$

Mt try: $$\Longleftrightarrow 3-(xy+yz+zx)+2\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge 4$$ or $$\sum_\text{cyc}\sqrt{(1-xy)(1-yz)}\ge\dfrac{1}{2}(1+xy+yz+zx)$$

River Li
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math110
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4 Answers4

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After squaring of the both sides we need to prove that $$2\sum\limits_{cyc}\sqrt{(1-xy)(1-xz)}\geq1+xy+xz+yz$$ or $$2\sum\limits_{cyc}\sqrt{((x+y+z)^2-4xy)((x+y+z)^2-4xz)}\geq(x+y+z)^2+4(xy+xz+yz)$$ or $$2\sum\limits_{cyc}\sqrt{((x-y)^2+z^2+2xz+2yz)((x-z)^2+y^2+2xy+2yz)}\geq\sum\limits_{cyc}(x^2+6xy)$$ and after using C-S it remains to prove that $$2\sum_{cyc}((x-y)(x-z)+xy+2x\sqrt{yz}+2xy)\geq\sum\limits_{cyc}(x^2+6xy)$$ or $$\sum_{cyc}(x^2-2xy+4x\sqrt{yz})\geq0$$ or $$\sum\limits_{cyc}\left(x^2-\sqrt{x^3y}-\sqrt{x^3z}+x\sqrt{yz}\right)+\sum_{cyc}\left(\sqrt{x^3y}+\sqrt{x^3z}-2xy\right)+3\sum_{cyc}x\sqrt{yz}\geq0$$ which is Schur and Muirhead.

Michael Rozenberg
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Lets minimise the LHS. WLOG let $x \ge y \ge z \ge 0$ - i.e. $x \in [y, 2], y \in [z, x], z \in [0, y]$.

The LHS is concave in all variables, hence its minima will happen when $x, y, z $ take extreme values in their allowable intervals. Along with the constraint, this leads to only three possibilities viz. $(x, y, z) \in \{(2, 0, 0), (1, 1, 0), (\frac23, \frac23, \frac23)\}$

In particular note that the minimum is given when $x=y=1, z=0$ (or permutations of it).

Macavity
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  • This problem is contest ,so maybe have other methods – math110 Aug 24 '16 at 04:54
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    @Macavity your solution is total wrong! There is a condition, otherwise we can check $(0,0,0)$, which does not work. – Michael Rozenberg Aug 24 '16 at 07:38
  • @MichaelRozenberg The condition / constraint is of course being used when finding the three possibilities. – Macavity Aug 24 '16 at 07:40
  • If so you can not use convexity because a variables depend of each other. – Michael Rozenberg Aug 24 '16 at 08:01
  • @MichaelRozenberg Think of it as unconstrained at first, then a closed convex compact domain imposed. The minimum has to be at the corners of the domain now by concavity of the objective function. Similar logic to your $u, v, w$ approach really. – Macavity Aug 24 '16 at 08:05
  • You can not say about convexity if you have a condition. OK. Why LS is a concave of $x$? Explain please. – Michael Rozenberg Aug 24 '16 at 08:13
  • By the way, the uvw's reasoning we use always after condition killing. – Michael Rozenberg Aug 24 '16 at 08:15
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    @MichaelRozenberg Check http://math.stackexchange.com/questions/748739/min-of-concave-symmetric-function-on-a-convex-set?rq=1 for an illustration. Couldn't find anything better on this site fast, but constrained convex optimization isn't exactly hard to find if you search. – Macavity Aug 24 '16 at 08:26
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    @Macavity you can not use here the convex optimization because there is a condition. I still think that your solution is total wrong. – Michael Rozenberg Sep 07 '16 at 02:53
  • @MichaelRozenberg At the risk of repeating everything, the condition is what defines the domain. Within the domain, which has to be closed and continuous (which it is here), convexity works. Simply put, no interior point of the domain can be the maximant due to the function being concave. Hence the result. I will see if a clearer eg is available somewhere. – Macavity Sep 07 '16 at 03:52
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Lets assume $z=\min\{x,y,z\}$ then note $\sqrt{1-xy}+\sqrt{1-yz}+\sqrt{1-xz}\geq\sqrt{1-\frac{(2-z)^2}{4}}+2\sqrt[4]{(1-yz)(1-xz)}=\sqrt{z-\frac{z^2}{4}}+2\sqrt[4]{1-z(2-z)+z^2xy}\geq\sqrt{z-\frac{z^2}{4}}+2\sqrt[4]{(1-z)^2+z^4}\geq 2$

gudduukar
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WLOG, assume that $x \ge y \ge z$.

We have $$\sqrt{1 - xy} \ge \sqrt{1 - (x + y)^2/4}.$$ Also, we have $$\sqrt{1 - yz} + \sqrt{1 - zx} \ge 1 + \sqrt{1 - yz - zx}.$$ (Note: Squaring both sides, it is just $2\sqrt{1 - yz - zx + z^2xy} \ge 2\sqrt{1 - yz - zx}$.)

Thus, it suffices to prove that $$\sqrt{1 - (x + y)^2/4} + 1 + \sqrt{1 - yz - zx} \ge 2$$ or $$\sqrt{1 - (2 - z)^2/4} + \sqrt{1 - (2 - z)z} \ge 1$$ or $$\frac12\sqrt{z(4 - z)} + (1 - z) \ge 1$$ which is true (using $z \le 2/3$).

We are done.

River Li
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  • -how did you come up with that solution, what were your thought process? – Philosophical Nature Jul 24 '22 at 10:31
  • @PhilosophicalNature First, we know $\sqrt{1 - u} + \sqrt{1 - v} \ge \sqrt{1} + \sqrt{1 - u - v}$ for $u, v \ge 0$ with $u + v \le 1$ (actually, one may use Karamata inequality for concave function $\sqrt u$). Second, the expression is minimized at $x = y = 1, z = 0$. It happens $\sqrt{1 - yz} + \sqrt{1 - zx} = \sqrt{1} + \sqrt{1 - yz - zx}$ at it. So I tried this estimate. – River Li Jul 24 '22 at 10:54