Find the minimum value of $P=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}$

Question

Let $x,y,z$ be real numbers such that $x>y>0, z>0$ and $xy+(x+y)z+z^2=1$. Find the minimum value of $$P=\frac{1}{4(x-y)^2}+\frac{1}{(x+z)^2}+\frac{1}{(y+z)^2}$$ I tried using some ways, but failed. Please give me an idea. Thank you.

Answer 1

notice that: $x - y = (x+z) - (y+z)$. So let $a = x+z$, and $b = y+z$, then:

$ab = 1$, and minimize: $P(a,b) = \dfrac{1}{4(a-b)^2} + \dfrac{1}{a^2} + \dfrac{1}{b^2} = a^2 + \dfrac{1}{a^2} + \dfrac{1}{4(a^2 - 2 + b^2)} = a^2 + \dfrac{1}{a^2} + \dfrac{1}{4\left(a^2 + \dfrac{1}{a^2} - 2\right)}= t + \dfrac{1}{4t - 8}$, with $t = a^2 + \dfrac{1}{a^2} \geq 2$.

So now consider $P(a,b) = f(t) = t + \dfrac{1}{4t-8}$ on $2 \leq t < \infty$.

We have: $f'(t) = 1 -\dfrac{4}{(4t-8)^2} = 1 - \dfrac{1}{4t^2 - 16t + 16} = 0 \iff 4t^2 - 16t + 16 = 1 \iff (2t-4)^2 = 1 \iff 2t - 4 = 1 \iff t = \dfrac{5}{2} \iff a = \sqrt{2}$, and then $b = \dfrac{1}{\sqrt{2}}$.

So: $f_{min} = f(\frac{5}{2}) = 3 = P_{min}$

Answer 2

From @DeepSea's above, suppose $b\equiv f(\,a\,),\,ab= 1\,\therefore\,{b}'= -\,\dfrac{b}{a},\,(\,a- 2\,b\,)(\,a- 2^{\,(\,\frac{1}{2}\,)}\,)\geqq 0$. $$a- b> 0 \tag{assume}$$ Let $$W(\,a\,)= \frac{1}{4(\,a- b\,)^{\,2}}+ \frac{1}{a^{\,2}}+ \frac{1}{b^{\,2}}- 3$$ $$\therefore\,{W}'(\,a\,)= -\,\frac{2}{a^{\,3}}+ \frac{{b}'- 1}{2(\,a- b\,)^{\,3}}- \frac{2\,{b}'}{b^{\,3}}$$ $$\therefore\,{W}'(\,a\,)= -\,\frac{2}{a^{\,3}}+ \frac{2}{ab^{\,2}}- \frac{-\,\dfrac{b}{a}- 1}{2(\,a- b\,)^{\,3}}$$ $$\therefore\,{W}'(\,a\,)= \frac{(\,a- 2\,b\,)(\,b- 2\,a\,)(\,a+ b\,)(\,2\,a^{\,2}- 3\,ab+ 2\,b^{\,2})}{2\,a^{\,3}b^{\,2}(\,a- b\,)^{\,2}}$$ $$\therefore\,(\,a- 2\,b\,){W}'(\,a\,)\geqq 0$$ $$\therefore\,(\,a- 2^{\,(\,\frac{1}{2}\,)}\,){W}'(\,a\,)\geqq 0$$ $$\therefore\,W(\,a\,)\geqq W(\,2^{\,(\,\frac{1}{2}\,)}\,)= 3$$ $$\because\,W(\,a\,)- W(\,2^{\,(\,\frac{1}{2}\,)}\,)= (\,a- 2^{\,(\,\frac{1}{2}\,)}\,){W}'(\,a\,) \tag{tangent equation}$$