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.
Best Answer
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