# Prove or disprove that the inequality is valid if $x,y,z,u$ are positive numbers and $x+y+z+u=2$.

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!

• If $c=1$ we can't have $c^2+b^2=2bc$ for all $b\in \{x,y,z,u\}$ unless $x=y=z=u=1$ but that makes $x+y+z+u=4.$ Instead,WLOG let $x=\max (x,y,z,u)$ and $y=\min (x,y,z,u)$ and consider $f(x)=\frac {x}{(x^2+1)^2}+\frac {y}{(y^2+1)^2},$ subject to $x+y$ being constant (so $\frac {dy}{dx}=-1$). Use calculus to show $f(x)$ is maximized only when $x=y.$ Commented Feb 1, 2023 at 23:47
• artofproblemsolving.com/community/c6h2629778p22735676 People are redirecting here and there because this is a very famous question. Finally I found a complete valid solution!😊
– user1034536
Commented Feb 2, 2023 at 8:32
• We may use $\frac{u^2}{(u^2+1)^2} \le \frac{4}{25}+\frac{48}{125}(u-1/2)$ for all $u \ge 3/25$. Commented Feb 2, 2023 at 12:04
• @youthdoo: If you have a “complete valid solution” then please post it here as an answer! Commented Feb 5, 2023 at 16:49
• @MartinR By "find", I actually meant find on the internet, not solving it myself.
– user1034536
Commented Feb 6, 2023 at 15:56

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 :

$$g(1/2-1/2*1/(x+1))=\frac{-8(14175x^{10}+151830x^{9}+725973x^{8}+2037080x^{7}+3707116x^{6}+4559360x^{5}+3823368x^{4}+2146160x^{3}+764600x^{2}+153600x+12800)}{(25(-5x^{2}-14x-10)^{2}(-5x^{2}-8x-4)^{2}(9x^{2}+18x+10)^{2})}< 0$$