 # Question about an inequality which seems right but not easy to prove The origin problem is as follo misurrosne 2022-06-24 Answered
Question about an inequality which seems right but not easy to prove
The origin problem is as follows:
let $a,b,c,d$ are positive real numbers,and $a+b+c+d=4$ prove:
$\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{d}+\frac{{d}^{2}}{a}\ge 4+\frac{1}{4}\left[\left(a-b{\right)}^{2}+\left(b-c{\right)}^{2}+\left(c-d{\right)}^{2}+\left(d-a{\right)}^{2}\right]$
The solution is easy enough, which is to plug $4=a+b+c+d$ into to $RHS$ and then move it to $LHS$, notice:
$\left[\frac{\left(a-b{\right)}^{2}}{b}+\frac{\left(b-c{\right)}^{2}}{c}+\frac{\left(c-d{\right)}^{2}}{d}+\frac{\left(d-a{\right)}^{2}}{a}\right]\left(a+b+c+d\right)\ge \left[\left(a-b{\right)}^{2}+\left(b-c{\right)}^{2}+\left(c-d{\right)}^{2}+\left(d-a{\right)}^{2}\right]$
which would finally lead to the proof of the problem.
However, when I tried to solve the problem, I applied the Lagrange identical equation to the $RHS$ of the inequality, then I left out some quadratic term which finally leads to the inequality:
$\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{d}+\frac{{d}^{2}}{a}\ge {a}^{2}+{b}^{2}+{c}^{2}+{d}^{2}$
Which seems quite right, but I'm not really sure if it is. However, numerical tests imply that it is right. But I'm still not sure.
That stuff is kind of like Chebyshev inequality but it can't be directly used here.
From another point of view, my question is whether the following strengthening　of the origin inequality is right or not:
let $a,b,c,d$ are positive real numbers,and $a+b+c+d=4$ then:
$\frac{{a}^{2}}{b}+\frac{{b}^{2}}{c}+\frac{{c}^{2}}{d}+\frac{{d}^{2}}{a}\ge 4+\frac{1}{4}\left[\left(a-b{\right)}^{2}+\left(b-c{\right)}^{2}+\left(c-d{\right)}^{2}+\left(d-a{\right)}^{2}+\left(a-c{\right)}^{2}+\left(b-d{\right)}^{2}\right]$
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$\sum _{cyc}\frac{{a}^{2}}{b}-4=\sum _{cyc}\left(\frac{{a}^{2}}{b}-2a+b\right)=\sum _{cyc}\frac{\left(a-b{\right)}^{2}}{b}\ge \frac{{\left(\sum _{cyc}|a-b|\right)}^{2}}{4}\ge \frac{1}{4}\sum _{cyc}\left(a-b{\right)}^{2}$