Determine the minimum value of: $P=\frac{\sqrt{x^2+xy+y^2}}{(x+y)^2+1}+\frac{\sqrt{y^2+yz+z^2}}{(y+z)^2+1}+\frac{\sqrt{z^2+zx+x^2}}{(z+x)^2+1}.$

Question

Given $x,y,z>0$ satisfy $x+y+z=\dfrac{3}{2}$.Determine the minimum value of: $$P=\frac{\sqrt{x^2+xy+y^2}}{(x+y)^2+1}+\frac{\sqrt{y^2+yz+z^2}}{(y+z)^2+1}+\frac{\sqrt{z^2+zx+x^2}}{(z+x)^2+1}.$$

I have tried:

$\bullet$ The minimum value is $\dfrac{3\sqrt{3}}{4}$ occur when $x=y=z=\dfrac{1}{2}$

$ \bullet P\ge \dfrac{\dfrac{\sqrt{3}}{2}(x+y)}{(x+y)^2+1}+\dfrac{\dfrac{\sqrt{3}}{2}(y+z)}{(y+z)^2+1}+\dfrac{\dfrac{\sqrt{3}}{2}(z+x)}{(z+x)^2+1}$

$\bullet$ So we need to prove $\dfrac{\dfrac{\sqrt{3}}{2}(x+y)}{(x+y)^2+1}\ge \dfrac{\sqrt{3}}{4}$ as well as prove $\dfrac{\dfrac{\sqrt{3}}{2}(y+z)}{(y+z)^2+1} \ge \dfrac{\sqrt{3}}{4}$ and $\dfrac{\dfrac{\sqrt{3}}{2}(z+x)}{(z+x)^2+1}\ge \dfrac{\sqrt{3}}{4}$

$\bullet$ Let $a=x+y$, the problem is $\dfrac{\dfrac{\sqrt{3}}{2}a}{a^2+1}\ge \dfrac{\sqrt{3}}{4}$ or $\dfrac{-(a-1)^2}{a^2+1}\ge0$ (which isn't true)

Pls help me with this problem and it would be nice if you could explain why my work is wrong.

Answer 1

The minimum does not exist.

Let $x\rightarrow\frac{3}{2}^-$ and $y=z\rightarrow0^+$.

Thus, we get a value $\frac{12}{13}.$

We'll prove that it's an infimum.

If it's given that $x\geq0$, $y\geq0$ and $z\geq0$ so for $(x,y,z)=\left(\frac{3}{2},0,0\right)$ we obtain a value $\frac{12}{13}$ again

and we'll prove that it's a minimal value.

Indeed, let $x=\max\{x,y,z\}$.

Thus, by C-S $$\sum_{cyc}\frac{\sqrt{x^2+xy+y^2}}{(x+y)^2+1}\geq\frac{x+\frac{y}{2}}{(x+y)^2+1}+\frac{x+\frac{z}{2}}{(x+z)^2+1}+\frac{\frac{5}{6}(y+z)}{(y+z)^2+1}\geq$$ $$\geq\frac{\left(x+\frac{y}{2}+x+\frac{z}{2}+\frac{5}{6}(y+z)\right)^2}{(x+y)^2\left(x+\frac{y}{2}\right)+(x+z)^2\left(x+\frac{z}{2}\right)+\frac{5}{6}(y+z)^3+2x+\frac{4}{3}(y+z)}=$$ $$=\tfrac{\frac{2}{3}(x+y+z)\left(2x+\frac{4}{3}(y+z)\right)^2}{(x+y)^2\left(x+\frac{y}{2}\right)+(x+z)^2\left(x+\frac{z}{2}\right)+\frac{5}{6}(y+z)^3+\frac{8}{9}\left(x+\frac{2}{3}(y+z)\right)(x+y+z)^2}$$ and it's enough to prove that: $$\tfrac{\frac{2}{3}(x+y+z)\left(2x+\frac{4}{3}(y+z)\right)^2}{(x+y)^2\left(x+\frac{y}{2}\right)+(x+z)^2\left(x+\frac{z}{2}\right)+\frac{5}{6}(y+z)^3+\frac{8}{9}\left(x+\frac{2}{3}(y+z)\right)(x+y+z)^2}\geq\frac{12}{13}$$ or $$303(y+z)x^2+4(43y^2+248yz+43z^2)x\geq(y+z)(104y^2-35yz+104z^2),$$ which is obvious