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For $x,y,z>0$ and $xy+yz+xz=1$, minimize $$P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}$$

My try: Let $xy=a; yz=b;zx=c \Rightarrow a+b+c=1$

$\Rightarrow x^2=\frac{ac}{b};y^2=\frac{ab}{c};z^2=\frac{bc}{a}$

Hence $$P=\sum \frac{1}{\frac{4ac}{b}-b+2}=\sum \frac{1}{\frac{4ac}{b}-b+2(a+b+c)}=\sum \frac{1}{\frac{4ac}{b}+2a+b+2c}=\sum \frac{b}{4ac+2ab+b^2+2bc}$$

$$P=\sum \frac{b}{(2a+b)(2c+b)} \geq \sum \frac{4b}{(2a+2b+2c)^2}=\sum \frac{b}{(a+b+c)^2}=1$$

And i need new method ?

  • 0
    I have one solution, but it's not feasibility . I can do it , where? – 2017-03-01
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    $\begin{align} \begin{cases} a+b+c=6(1) \\ 0\leq a,b,c\leq4(2) \end{cases} \end{align}$ $(1)\Rightarrow \begin{align} \begin{cases} b+c=(6-a) \\b^2+c^2+bc=(6-a)^2-bc \end{cases} \end{align}$ – 2017-03-01
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    $\begin{align} \begin{cases} a+b+c=6(1) \\ 0\leq a,b,c\leq4(2) \end{cases} \end{align}$ $(1)\Rightarrow \begin{align} \begin{cases} b+c=(6-a) \\b^2+c^2+bc=(6-a)^2-bc \end{cases} \end{align}$ $P=a^2+(b^2+c^2+bc)+a(b+c)=a^2+[(6-a)^2-bc]+a(6-a)$ $P=(a^2-12a+36)-bc=(a-6)^2-bc$ $(2)\Rightarrow bc\leq 0 \Rightarrow P\geq (a-6)^2$ When $bc=0 \Rightarrow [{\begin{matrix}b=0\\c=0\end{matrix}}(3)$ $(1)and (3) \Rightarrow 2\leq a\leq 4(4)$ $P_{max} => |a-6|$ max satisfy $(4) $ $=>a=2$ from (1) and (3) $=> b=c=4$ $=>P_{max}(a,b,c)=P(4;2;0)=28$ – 2017-03-01
  • 0
    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/54562/discussion-between-s-c-b-and-word-shallow). – 2017-03-01

2 Answers 2

1

If $x=y=z=\frac{1}{\sqrt3}$ we get $P=1$.

We'll prove that it's a minimal value. Indeed, we need to prove that $$\sum_{cyc}\frac{1}{4x^2-yz+2}\geq1$$ or $$\sum_{cyc}\left(\frac{1}{4x^2-yz+2}-\frac{1}{3}\right)\geq0$$ or $$\sum_{cyc}\frac{1-4x^2+yz}{4x^2-yz+2}\geq0$$ or $$\sum_{cyc}\frac{xy+xz+2yz-4x^2}{4x^2-yz+2}\geq0$$ or $$\sum_{cyc}\frac{(z-x)(2x+y)-(x-y)(2x+z)}{4x^2-yz+2}\geq0$$ or $$\sum_{cyc}(x-y)\left(\frac{2y+z}{4y^2-xz+2}-\frac{2x+z}{4x^2-yz+2}\right)\geq0$$ or $$\sum_{cyc}(x-y)^2(z^2+2xy+2)(4z^2-xy+2)\geq0$$ Done!

1

You proved that $P \ge 1$.

And for $x=y=z=\frac{1}{\sqrt{3}}$ you have $$P=\frac{1}{4x^{2}-yz+2}+\frac{1}{4y^{2}-zx+2}+\frac{1}{4z^{2}-xy+2}= \frac{1}{3} + \frac{1}{3}+\frac{1}{3}=1.$$ Hence the minimum of $P$ for positive values of $x,y,z$ is one.