Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\geq\sqrt{\frac{24(a^2+b^2+c^2+d^2)}{ab+ac+bc+ad+bd+cd}}$$

This inequality is similar to the following inequality of three variables.

Let $a$, $b$ and $c$ be positive numbers. Prove that: $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq3\sqrt{\frac{a^2+b^2+c^2}{ab+ac+bc}},$$ which after squaring of the both sides is equivalent to $$\sum_{cyc}\left(\frac{a^3}{b}+\frac{a^3c}{b^2}+2ab+\frac{2a^2b}{c}+\frac{a^2c}{b}-7a^2\right)\geq0,$$ which is just AM-GM.

This method does not help for the starting inequality.

A tried also another ways, but without any success.

For example, by Holder $$\left(\sum_{cyc}\frac{a}{b}\right)^2\sum_{cyc}a^4b^2\geq\left(\sum_{cyc}a^2\right)^3.$$ Hence, it remains to prove that $$(a^2+b^2+c^2+d^2)^2(ab+ac+bc+ad+bd+cd)\geq24(a^4b^2+b^4c^2+c^4d^2+d^4a^2),$$ which is wrong for $c=d\rightarrow0^+$.

  • $\begingroup$ after squaring it can be proven by the substitution $$b=a+u,c=a+u+v,d=a+u+v+w$$ $\endgroup$ Apr 8, 2017 at 12:45
  • $\begingroup$ Sonnhard, it's hard for me. $\endgroup$ Apr 8, 2017 at 12:47
  • $\begingroup$ yes for me too but it can be solved i'm here in Vienna for a short visit $\endgroup$ Apr 8, 2017 at 12:48
  • $\begingroup$ @Dr.SonnhardGraubner Would you mind to work out your first comment and give an answer to the OP? $\endgroup$
    – Hanno
    Dec 8, 2017 at 11:24

1 Answer 1


This is an outline of a proof.

It suffices to prove that $$a^2b^2c^2d^2(ab+bc+cd+da+ac+bd)\Big[\Big(\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a}\Big)^2 - \frac{24(a^2+b^2+c^2+d^2)}{ab+bc+cd+da+ac+bd}\Big] \ge 0.$$ Denote LHS by $f(a,b,c,d)$ which is a homogeneous polynomial.

WLOG, let $d = \min(a,b,c,d)$. We use the Buffalo Way. Let $c = d + s, \ b = d + t, \ a = d + r; \ s, t, r \ge 0.$ We have $$f(d+r, d+t, d+s, d) = a_8d^8 + a_7d^7 + a_6d^6 + a_5d^5 + a_4d^4 + a_3d^3 + a_2 d^2 + a_1d + a_0$$ where \begin{align} a_8 &= 24 r^2+16 r s-32 r t+24 s^2-32 s t+24 t^2, \\ a_7 &= 24 r^3+104 r^2 s-16 r^2 t+104 r s^2-144 r s t+32 r t^2+24 s^3+32 s^2 t-16 s t^2+24 t^3, \\ a_6 &= 6 r^4+96 r^3 s-12 r^3 t+220 r^2 s^2-60 r^2 s t+50 r^2 t^2+96 r s^3+60 r s^2 t-116 r s t^2+60 r t^3+6 s^4+60 s^3 t+50 s^2 t^2-12 s t^3+6 t^4, \\ a_5 &= 3 r^5+23 r^4 s-19 r^4 t+162 r^3 s^2-8 r^3 s t+23 r^3 t^2+162 r^2 s^3+174 r^2 s^2 t-51 r^2 s t^2+79 r^2 t^3+23 r s^4+168 r s^3 t+45 r s^2 t^2-34 r s t^3+17 r t^4+3 s^5+17 s^4 t+79 s^3 t^2+23 s^2 t^3-19 s t^4+3 t^5,\\ a_4 &= 7 r^5 s+r^5 t+34 r^4 s^2-37 r^4 s t-12 r^4 t^2+106 r^3 s^3+122 r^3 s^2 t+r^3 s t^2+43 r^3 t^3+34 r^2 s^4+234 r^2 s^3 t+60 r^2 s^2 t^2+29 r^2 s t^3+20 r^2 t^4+7 r s^5+49 r s^4 t+157 r s^3 t^2+33 r s^2 t^3-39 r s t^4+7 r t^5+7 s^5 t+20 s^4 t^2+43 s^3 t^3-12 s^2 t^4+s t^5,\\ a_3 &= 5 r^5 s^2+2 r^5 s t+19 r^4 s^3-5 r^4 s^2 t-30 r^4 s t^2+2 r^4 t^3+19 r^3 s^4+136 r^3 s^3 t+34 r^3 s^2 t^2+48 r^3 s t^3+9 r^3 t^4+5 r^2 s^5+59 r^2 s^4 t+130 r^2 s^3 t^2+38 r^2 s^2 t^3-15 r^2 s t^4+5 r^2 t^5+16 r s^5 t+43 r s^4 t^2+80 r s^3 t^3-24 r s^2 t^4+2 r s t^5+5 s^5 t^2+9 s^4 t^3+2 s^3 t^4,\\ a_2 &= r^5 s^3+r^5 s^2 t+2 r^4 s^4+15 r^4 s^3 t-16 r^4 s^2 t^2+2 r^4 s t^3+r^3 s^5+29 r^3 s^4 t+50 r^3 s^3 t^2+24 r^3 s^2 t^3+5 r^3 s t^4+r^3 t^5+11 r^2 s^5 t+34 r^2 s^4 t^2+44 r^2 s^3 t^3-10 r^2 s^2 t^4+r^2 s t^5+11 r s^5 t^2+17 r s^4 t^3+4 r s^3 t^4+s^5 t^3, \\ a_1 &= 2 r^4 s^4 t+2 r^4 s^3 t^2+2 r^3 s^5 t+11 r^3 s^4 t^2+6 r^3 s^3 t^3+2 r^3 s^2 t^4+7 r^2 s^5 t^2+9 r^2 s^4 t^3+2 r^2 s^3 t^4+2 r s^5 t^3, \\ a_0 &= r^3 s^5 t^2+r^3 s^4 t^3+r^2 s^5 t^3. \end{align} It suffices to prove that $a_i\ge 0$ for $i=0, 1, \cdots, 8.$

Maybe someone can give a nice proof that $a_i\ge 0$ for $i=0, 1, \cdots, 8.$

  • $\begingroup$ Did you really prove that all $a_i\geq0?$ I see that $a_8\geq0$, $a_7\geq0$, $a_1\geq0$ and $a_0\geq0$. $\endgroup$ Jul 18, 2019 at 15:04
  • 1
    $\begingroup$ I can write $a_8, a_7, a_6, a_5, a_4, a_1, a_0$ in the form of $z^TQz$ where $Q$ is positive semi-definite rational matrix. However, for $a_2, a_3$, I can write it in the form of $z^TAz$ where $A$ is semi-definite real matrix (round-off error). Based on this, I believe $a_2\ge 0, a_3\ge 0$. A rigorous proof is to find rational SOS. $\endgroup$
    – River Li
    Jul 18, 2019 at 15:30

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