# How find this minimun of this $\frac{xy}{x^5+xy+y^5}+\frac{yz}{y^5+yz+z^5}+\frac{xz}{x^5+xz+z^5}$

let $x,y,z>0$ and such $$x+y+z=1$$ Find minimum of the $$\dfrac{xy}{x^5+xy+y^5}+\dfrac{yz}{y^5+yz+z^5}+\dfrac{xz}{x^5+xz+z^5}$$

I have find this maximum

note $$\dfrac{xy}{x^5+xy+y^5}\le -243\dfrac{x+y}{841}+\dfrac{23031}{1682}$$ see http://www.wolframalpha.com/input/?i=xy%2F%28x%5E5%2Bxy%2By%5E5%29%2B243*%28x%2By%29%2F841-23031%2F1682

But for minimum value,I can't,(maybe use AM-GM) Thank you,

• I can't see it being anything except $81/29.$ If all three have to be positive, then there will be trade-offs if $x \neq y \neq z$ because of the cyclic symmetry of the function to minimize. So $x = y = z = 1/3.$ But I can't prove it. – John Jan 8 '14 at 4:14
• For $g(x,y)=xy/(x^5+xy+y^5)$ if $\partial_x^2 g(x,y) < 0$ for $0<x,y\leq 1$, which some graphing may suggest, then I think there is an argument using Lagrange multipliers that $x=y=z$ is where the min occurs. But it is too late now... – abnry Jan 8 '14 at 4:41
• Lagrange multipliers looks hairy. I'm guessing there's a trick. – John Jan 8 '14 at 5:09
• @John Lagrange multipliers work if $g_x(x,y)$ is one-to-one for all $y<1$. All the symmetries work out otherwise in just a fine manner (with some argument). – abnry Jan 8 '14 at 5:48

The max is $\dfrac{81}{29}$ and min is $\dfrac{.25}{2*(0.5)^5+.25}$, a simple method is let two varies equal$(x=y)$ and have a function of $x$, see graphic below:
edit :to prove max, consider $xy \le \dfrac{x^2+y^2}{2}, x^5+y^5 \ge 2(\dfrac{x^2+y^2}{2})^{\frac{5}{2}},\dfrac{xy}{x^5+xy+y^5}\le \dfrac{x^2+y^2}{x^2+y^2+4(\dfrac{x^2+y^2}{2})^{\frac{5}{2}}}=\dfrac{p}{p+4(\dfrac{p}{2})^{\frac{5}{2}}}=f(p)$
$f(p)$ is concave and mono decreasing function and $x^2+y^2+z^2 \ge \dfrac{1}{3}$ ,so we can quickly know the max will be got when $x=y=z$
• Hello,the max is how prove it? becasue I think $$\dfrac{xy}{x^5+xy+y^5}\le\dfrac{23031}{1682}-\dfrac{243}{841}(x+y)$$ is not easy prove – math110 Jan 8 '14 at 7:42
• @math110 this value is wrong!$\dfrac{xy}{x^5+xy+y^5}< 1 \implies$ max $<3$,but $\dfrac{23031}{1682}>13,x+y<1 \to$ Max $>12$ – chenbai Jan 8 '14 at 7:51