Prove or disprove that the inequality $$ \dfrac{x^2}{\left(x^2+1\right)^2}+\dfrac{y^2}{\left(y^2+1\right)^2}+\dfrac{z^2}{\left(z^2+1\right)^2}+\dfrac{u^2}{\left(u^2+1\right)^2} \leq \dfrac{16}{25}$$ is valid if $x,y,z,u$ are positive numbers and $x+y+z+u=2.$ What do I do? First I use this $$c^2+b^2 \geq 2bc.$$ If $$c^2+b^2 \geq 2bc,$$ then $$\dfrac{1}{c^2+b^2} \leq \dfrac{1}{2bc}.$$ So we have $$ \dfrac{x^2}{\left(x^2+1\right)^2}+\dfrac{y^2}{\left(y^2+1\right)^2}+\dfrac{z^2}{\left(z^2+1\right)^{2}}+\dfrac{u^2}{\left(u^2+1\right)^{2}} \leq \dfrac{x^2}{(2x)^{2}}+\dfrac{y^2}{(2y)^2}+\dfrac{z^2}{(2z)^2}+\dfrac{u^2}{(2u)^2}=\dfrac{x^2}{4x^2}+\dfrac{y^2}{4y^2}+\dfrac{z^2}{4z^2}+\dfrac{u^2}{4u^2}=\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}=1 \geq \dfrac{16}{25}.$$
But as I understand $1$ is just a maximum, so the initial inequality can still be less than $\dfrac{16}{25}$. Any hint would help a lot! Thanks in advance!
Partial answer :
As user @alet show the inequality as the variable are not in $[0.5,2/3]$ I complete it :
Let $a,c,d\in[0.5,1/\sqrt{3}]$ and $b\in[0,0.5]$ such that $a+b+c+d=2$ and $a\geq d\ge c\ge b$ then we have :
$$af\left(a\right)+b\left(b\right)+cf\left(c\right)+df\left(d\right)\leq \left(2-b\right)f\left(\frac{a^{2}+c^{2}+d^{2}}{2-b}\right)+bf\left(b\right)-\frac{16}{25}\leq 0\leq 16/25$$
where :
$$f\left(x\right)=\frac{x}{\left(x^{2}+1\right)^{2}}$$
The LHS is just weighted Jensen's inequality because for $x\in[0,1]$ :
$$f''(x)=\frac{12x\left(x^{2}-1\right)}{\left(x^{2}+1\right)^{4}}\leq 0$$
For the LHS we have $b=x$:
$$g(x)=\left(2-x\right)f\left(\frac{a^{2}+c^{2}+d^{2}}{2-x}\right)+xf\left(x\right)-\frac{16}{25}\leq \left(2-x\right)f\left(\frac{1.25-x+\left(0.5-x\right)^{2}}{2-x}\right)+xf\left(x\right)-\frac{16}{25}$$
Or :
$g(1/2-1/2*1/(x+1))=\frac{-8(14175x^{10}+151830x^{9}+725973x^{8}+2037080x^{7}+3707116x^{6}+4559360x^{5}+3823368x^{4}+2146160x^{3}+764600x^{2}+153600x+12800)}{(25(-5x^{2}-14x-10)^{2}(-5x^{2}-8x-4)^{2}(9x^{2}+18x+10)^{2})}< 0$