# How to prove this inequality $\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$

How to prove this inequality

$$\sqrt{\frac{ab+bc+cd+da+ac+bd}{6}}\geq \sqrt[3]{{\frac{abc+bcd+cda+dab}{4}}}$$

for $$a,b,c,d\gt0$$?

Thanks

• Are there any constraints on a,b,c,d? – NoChance Sep 18 '12 at 8:38
• @NoChance: I believe that the variables should be positive. The answers so far assume they are, so I have added that condition. – robjohn Jan 26 '20 at 23:58
• @tianzhidaosunyouyu: I have added the condition that $a,b,c,d\gt0$. If that should not be added, feel free to revert my edit. – robjohn Jan 27 '20 at 0:00
• @robjohn, thanks for the clarification. – NoChance Jan 27 '20 at 1:12

This is a special case of the so-called Maclaurin's inequality.

There is also another take on this. The two sides of this inequality are called "symmetric monomial means". The LHS is denoted $$\mathfrak{M}_{11}$$ and the RHS is denoted $$\mathfrak{M}_{111}$$, so the inequality claims that $$\mathfrak{M}_{11}\ge \mathfrak{M}_{111}$$. The research on the inequalities between monomial means still has some interesting unsolved problems.

EDIT

IF you want to prove this directly there is an interesting substitution which is good for those inequalities. If $$LHS \ge RHS$$ is your inequality it is equivalent to $$LHS^6-RHS^6\ge 0$$. The last one is a polynomial in a,b,c,d. Now make the following substituion: $${a=x + y + z + t, b = y + z + t, c = z + t, d = t}$$. The resulting degree 6 polynomial will have 70 coefficients and all of them will be positive. Of course carrying out this strategy by hand is awkward, but here is what Mathematica spilled out at the end:$$(3 t^4 x^2)/16 + (t^3 x^3)/8 + 1/4 t^4 x y + 1/2 t^3 x^2 y + 1/8 t^2 x^3 y + (t^4 y^2)/4 + 3/4 t^3 x y^2 + 11/24 t^2 x^2 y^2 + 1/24 t x^3 y^2 + (t^3 y^3)/2 + 2/3 t^2 x y^3 + 1/6 t x^2 y^3 + ( x^3 y^3)/216 + (t^2 y^4)/3 + 5/24 t x y^4 + (x^2 y^4)/72 + ( t y^5)/12 + (x y^5)/72 + y^6/216 + 1/8 t^4 x z + 5/8 t^3 x^2 z + 1/4 t^2 x^3 z + 1/4 t^4 y z + 5/4 t^3 x y z + 29/24 t^2 x^2 y z + 1/6 t x^3 y z + 5/4 t^3 y^2 z + 17/8 t^2 x y^2 z + 17/24 t x^2 y^2 z + 1/36 x^3 y^2 z + 17/12 t^2 y^3 z + 13/12 t x y^3 z + 1/9 x^2 y^3 z + 13/24 t y^4 z + 5/36 x y^4 z + ( y^5 z)/18 + (3 t^4 z^2)/16 + 5/8 t^3 x z^2 + 5/6 t^2 x^2 z^2 + 1/6 t x^3 z^2 + 5/4 t^3 y z^2 + 7/3 t^2 x y z^2 + t x^2 y z^2 + 1/18 x^3 y z^2 + 7/3 t^2 y^2 z^2 + 2 t x y^2 z^2 + 37/144 x^2 y^2 z^2 + 4/3 t y^3 z^2 + 29/72 x y^3 z^2 + ( 29 y^4 z^2)/144 + (5 t^3 z^3)/8 + t^2 x z^3 + 1/2 t x^2 z^3 + ( x^3 z^3)/27 + 2 t^2 y z^3 + 7/4 t x y z^3 + 19/72 x^2 y z^3 + 7/4 t y^2 z^3 + 41/72 x y^2 z^3 + (41 y^3 z^3)/108 + (3 t^2 z^4)/4 + 5/8 t x z^4 + (5 x^2 z^4)/48 + 5/4 t y z^4 + 5/12 x y z^4 + ( 5 y^2 z^4)/12 + (3 t z^5)/8 + (x z^5)/8 + (y z^5)/4 + z^6/16$$

This is obviously positive so the inequality is proved.

• I can't believe you typed all this on a keyboard! Good job indeed! – NoChance Sep 18 '12 at 6:02
• I didn't type it. I copied it from Mathematica. – ivan Sep 18 '12 at 6:06
• Even more clever! +1, well deserved! – NoChance Sep 18 '12 at 6:08
• The link to the PDF file does not work (“You don't have permission to access this resource”). – Martin R Jan 21 '20 at 10:36

I remember seeing this problem in some book a couple of years ago. My attention was drawn by the proof of this inequality, which apparently originates from the '70 GDR mathematical olympiad. The solution involves nothing more than AM-GM (I assume $a,b,c,d$ are nonnegative here!), but some algebraic transformations are pretty insane, so stay calm $\ddot\smile$

$$\begin{split} \quad &\sqrt{\dfrac{ab + ac + ad + bc + bd + cd}{6}} = \\ &\sqrt{\dfrac{(ab+cd)/2 + (ac+bd)/2 + (ad+bc)/2}{3}} \geq \quad \text{// AM-GM applied here} \\ &\sqrt[6]{\dfrac{(ab+cd)(ac+bd)(ad+bc)}{8}} = \\ &\sqrt[6]{\dfrac{a^3bcd + ab^3cd + abc^3d + abcd^3}{8} + \dfrac{a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2}{8}} = \\ &\sqrt[6]{\dfrac{\left(\dfrac{a^2 + b^2}{2} + \dfrac{b^2 + c^2}{2} + \dfrac{c^2 + d^2}{2} + \dfrac{d^2 + a^2}{2}\right)abcd + \dfrac{a^2 + c^2}{2}b^2d^2 + \dfrac{b^2 + d^2}{2}a^2c^2}{8} + } \\ &\hspace{120pt} \overline{+ \dfrac{a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2}{16}} \geq \quad \text{// and here}\\ &\sqrt[6]{\dfrac{a^2b^2c^2 + a^2b^2d^2 + a^2c^2d^2 + b^2c^2d^2}{16} +} \\ &\hspace{120pt} \overline{+ \dfrac{2(a^2b^2cd + ab^2c^2d + abc^2d^2 + a^2bcd^2 + ab^2cd^2 + a^2bc^2d)}{16}} = \\ &\sqrt[6]{\left(\dfrac{abc + abd + acd + bcd}{4}\right)^2} = \sqrt[3]{\dfrac{abc + abd + acd + bcd}{4}} _{\square} \end{split}$$

Side note: if anyone knows how to improve the formatting (make the multi-line 6th root look smoother), feel free to edit this. :) I also used \overline to make the two long root expressions a bit more readable, I hope it renders properly.

Here is my proof for positive variables from 1979.

Let $a+b+c+d=4u$, $ab+ac+bc+ad+bd+cd=6v^2$, $abc+abd+acd+bcd=4w^3$ and $abcd=t^4$, where $v>0$.

Hence, $a$, $b$, $c$ and $d$ are positive roots of the following equation:

$(x-a)(x-b)(x-c)(x-d)=0$ or $x^4-4ux^3+6v^2x^2-4w^3x+t^4=0$.

Hence, by Rolle the equation $(x^4-4ux^3+6v^2x^2-4w^3x+t^4)'=0$ or $x^3-3ux^2+3v^2x-w^3=0$ has three positive roots.

Since we can replace $x$ at $\frac{1}{x}$, we get that the equation $w^3x^3-3v^2x^2+3ux-1=0$ has three positive roots.

Now by the Rolle's theorem again we get that the equation

$(w^3x^3-3v^2x^2+3ux-1)'=0$ or $w^3x^2-2v^2x+u=0$ has two real roots, which says that $v^4\geq uw^3$.

By the Rolle's theorem again the equation $(x^3-3ux^2+3v^2x-w^3)'=0$

or $x^2-2ux+v^2=0$ has two positive roots, which says that $u^2\geq v^2$ or $u\geq v$.

The last inequality is also $\sum\limits_{sym}(a-b)^2\geq0$.

Id est, $v^4\geq uw^3\geq vw^3$, which gives $v\geq w$ and we are done!

• thanks for your wonderful proof! – vidyarthi Jan 20 '17 at 16:52

Definitions

Suppose $$p_n(x)$$ is a degree $$n$$ polynomial with real roots, $$\left\{-r_{n,k}\right\}_{k=1}^n$$, where $$0\lt r_{n,k}\lt r_{n,k+1}$$. Define $$a_k$$ as follows: \begin{align} p_n(x) &=\prod_{k=1}^n(x+r_{n,k})\tag1\\ &=\sum_{k=0}^n\binom{n}{k}a_kx^{n-k}\tag2 \end{align} That is, $$\binom{n}{k}a_k$$ is the degree $$k$$ elementary symmetric polynomial on $$\{r_{n,j}\}_{j=1}^n$$.

Then, define \begin{align} p_{n-1}(x) &=\frac1n\,p_n'(x)\tag3\\[3pt] &=\sum_{k=0}^{n-1}\binom{n-1}{k}a_kx^{n-1-k}\tag4\\ &=\prod_{k=1}^{n-1}(x+r_{n-1,k})\tag5 \end{align} That is, $$p_{n-1}(x)$$ is a degree $$n-1$$ polynomial. The Mean Value Theorem says that $$p_{n-1}(x)$$ has real roots, $$\left\{-r_{n-1,k}\right\}_{k=1}^{n-1}$$, where $$r_{n,k}\lt r_{n-1,k}\lt r_{n,k+1}$$. Furthermore, we have that $$\binom{n-1}{k}a_k$$ is the degree $$k$$ elementary symmetric polynomial on $$\{r_{n-1,j}\}_{j=1}^{n-1}$$.

Extension by Induction

Induction gives that for any $$m\le n$$, \begin{align} p_m(x) &=\sum_{k=0}^m\binom{m}{k}a_kx^{m-k}\tag6\\ &=\prod_{k=1}^m(x+r_{m,k})\tag7 \end{align} where $$0\lt r_{m,k}\lt r_{m,k+1}$$ and $$\binom{m}{k}a_k$$ is the degree $$k$$ elementary symmetric polynomial on $$\{r_{m,j}\}_{j=1}^m$$.

In particular, the degree $$m-1$$ elementary symmetric polynomial on $$\{r_{m,j}\}_{j=1}^m$$ is $$ma_{m-1}=\left(\sum_{k=1}^m\frac1{r_{m,k}}\right)\prod_{k=1}^mr_{m,k}\tag8$$ and the degree $$m$$ elementary symmetric polynomial on $$\{r_{m,j}\}_{j=1}^m$$ is $$a_m=\prod_{k=1}^mr_{m,k}\tag9$$

Inequalities for the Elementary Symmetric Functions

The AM-GM inequality gives \begin{align} a_{m-1} &=\frac1m\left(\sum_{k=1}^m\frac1{r_{m,k}}\right)\prod_{k=1}^mr_{m,k}\tag{10}\\ &\ge\left(\prod_{k=1}^mr_{m,k}^{-1/m}\right)\prod_{k=1}^mr_{m,k}\tag{11}\\ &=\prod_{k=1}^mr_{m,k}^{\frac{m-1}m}\tag{12}\\[6pt] &=a_m^{\frac{m-1}m}\tag{13} \end{align} Therefore, we have that $$a_{m-1}^{\ \ \frac1{m-1}}\ge a_m^{\ \frac1m}\tag{14}$$

Apply the case $$m=3$$ of $$(14)$$ and we get $$a_2^{1/2}\ge a_3^{1/3}$$. Setting $$n=4$$, we get $$\left(\frac{ab+bc+ca+ad+bd+cd}6\right)^{1/2}\ge\left(\frac{abc+bcd+cda+dab}4\right)^{1/3}\tag{15}$$