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!

  • 3
    $\begingroup$ 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 $x=y.$ $\endgroup$ Feb 1 at 23:47
  • 1
    $\begingroup$ artofproblemsolving.com/community/c6h2629778p22735676 People are redirecting here and there because this is a very famous question. Finally I found a complete valid solution!😊 $\endgroup$
    – user1034536
    Feb 2 at 8:32
  • $\begingroup$ We may use $\frac{u^2}{(u^2+1)^2} \le \frac{4}{25}+\frac{48}{125}(u-1/2)$ for all $u \ge 3/25$. $\endgroup$
    – River Li
    Feb 2 at 12:04
  • $\begingroup$ @youthdoo: If you have a “complete valid solution” then please post it here as an answer! $\endgroup$
    – Martin R
    Feb 5 at 16:49
  • $\begingroup$ @MartinR By "find", I actually meant find on the internet, not solving it myself. $\endgroup$
    – user1034536
    Feb 6 at 15:56

1 Answer 1


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 :


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 :

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .