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let $x,y,z>0$ and such $$x^2+y^2+z^2+2xyz=1$$ show that $$xy+yz+xz\le 2xyz+\dfrac{1}{2}$$

My try: since $$1=x^2+y^2+z^2+2xyz\ge xy+yz+xz+2xyz$$ then $$xy+yz+xz\le 1-2xyz$$ so we only prove follow this $$1-2xyz\le 2xyz+\dfrac{1}{2}$$ $$\Longleftrightarrow xyz\ge\dfrac{1}{4}$$ But this is not true,so How prove my inequality? Thank you

  • At your first step the quantity decreases when the square sum is replaced by the sum of adjacent products. So at the "we only need prove" step it may be the new inequality doesn't hold. Another thing: the last iff should be iff $xyz \ge 1/8$ (maybe something can be done with this milder inequality). – coffeemath Jan 25 '14 at 17:14
  • Note that my "answer" below had an error (if you looked at it), and that the constraint does not imply $xyz \ge 1/8$ so there must be some other method to prove the inequality. – coffeemath Jan 25 '14 at 19:42

4 Answers4

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We first we first notice that there are always two of the three numbers, both greater than $\dfrac{1}{2}$,because of symmetry,we may assume that $x,y\le\dfrac{1}{2},$ or $x,y\ge \dfrac{1}{2}$ and then $$(2x-1)(2y-1)\ge 0\Longleftrightarrow x+y-2xy\le\dfrac{1}{2}$$ on the other hand, $$1=x^2+y^2+z^2+2xyz\ge 2xy+z^2+2xyz$$ then $$2xy(1+z)\le 1-z^2\Longrightarrow 2xy\le 1-z$$ we only have to multiply side by side the inequality from above $$x+y-2xy\le\dfrac{1}{2},z\le 1-2xy$$ then $$xz+yz-2xyz\le\dfrac{1}{2}-xy\Longleftrightarrow xy+xz+yz\le\dfrac{1}{2}+2xyz$$

math110
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    Nice +1. You just need that any two among $\frac12-x, \frac12-y, \frac12-z$ have the same sign, which is obvious. – Macavity Jan 26 '14 at 05:48
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Edit: I found the following has a false claim that $xyz\ge1/8$ follows from the constraint. I'll leave it up for now (will delete if asked). At the end I give an example where $xyz<1/8.$

Your proof can be finished, given you correct the final inequality. Note the equivalent inequalities $$1-2xyz \le 2xyz+1/2, \\ 1/2 \le 4xyz,\\ xyz\ge 1/8.$$ Now from the constraint $x^2+y^2+z^2+2xyz=1$ and the objective to minimize $xyz$ one can use lagrange multipliers to conclude that, for positive $x,y,z$, one has $x=y=z$ at any critical point. With each of these put equal to $t$ we get $2t^3+3t^2-1=(t+1)^2(2t-1)=0$ giving $t=1/2$ [can't use $t=-1$ here]where $xyz=1/8.$

A bit more work is needed to verify this is indeed the minimum of $xyz$ given the constraint.

Edit: A bit more indeed! It must be these internal critical points don't give the global minimum of $xyz$, since $x=y=3/5,z=7/25$ satisfies the constraint $x^2+y^2+z^2+2xyz=1$ and yet $xyz=63/625=0.1008<1/8=0.125.$

coffeemath
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    In fact $1 = x^2+y^2+z^2+2xyz \ge 3(xyz)^{2/3}+2(xyz) \implies xyz \le \frac18$, so we always have this problem! You may have found the maximum. – Macavity Jan 26 '14 at 04:47
  • @Macavity I looked at replacing the constraints to nonnegative $x,y,z$ and tried LaGrange, but apart from some nice factors implying equality of the variables, there were other unwieldy ones which I couldn't show implied anything simple. – coffeemath Jan 26 '14 at 05:42
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    Know the feeling ;) math110's solution looks good. – Macavity Jan 26 '14 at 05:51
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Let $x=\frac{a}{\sqrt{(a+b)(a+c)}}$ and $y=\frac{b}{\sqrt{(a+b)(b+c)}}$, where $a$, $b$ and $c$ are positives.

Hence, $z=\frac{c}{\sqrt{(a+c)(b+c)}}$ and we need to prove that $$\sum_{cyc}\frac{ab}{(a+b)\sqrt{(a+c)(b+c)}}\leq\frac{2abc}{\prod\limits_{cyc}(a+b)}+\frac{1}{2}$$ or $$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq4abc+\prod_{cyc}(a+b)$$ or

$$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq\sum_{cyc}(a^2b+a^2c+2abc),$$ which is AM-GM: $$2\sum_{cyc}ab\sqrt{(a+c)(b+c)}\leq\sum_{cyc}ab(a+b+2c)=\sum_{cyc}(a^2b+a^2c+2abc).$$ Done!

Michael Rozenberg
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Here's a solution using the Lagrange multiplier method. We are looking for extrema of $f(x,y,z) = xy+yz+xz-2xyz$ given that $x,y,z \geq 0$ and $x^2+y^2+z^2+2xyz=1$. On the boundary (where one variable is $0$) we have $f \in [0,\tfrac{1}{2}]$. Now assume $x,y,z>0$. Define $s=x+y+z$. Then the Lagrange multiplier method shows that at an extremum of $f$ we must have $$\begin{eqnarray} (y-z)(x^2-2sx+s) &=& 0\\ (x-z)(y^2-2sy+s) &=& 0\\ (x-y)(z^2-2sz+s) &=& 0 \end{eqnarray} $$

Consider the first equality. This implies $y=z$ or $x > \tfrac{1}{2}$. Since not all of $x,y,z$ can be greater than $\tfrac{1}{2}$ we conclude that at least two of these variables must be equal. Now $f(x,x,1-2x^2)$ has a maximum of $\tfrac{1}{2}$ on the interval $(0,\sqrt{1/2})$ at $x=\tfrac{1}{2}$ (and a local minimum at $x=(1+\sqrt{17})/8$).

WimC
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