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

Question

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

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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 ?

Answer 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!

Answer 2

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.