# Dorin Marghidanu's Light Elegance in Four Variables

### Source

### Problem

### Solution 1

Let $-a+b+c+d=2x,\,$ $a-b+c+d=2y,\,$ $a+b-c+d=2z,\,$ $a+b+c-d=2t.\,$ Then $a+b+c+d=x+y+z+t\,$ and the required inequality becomes

$4(x^2+y^2+z^2+t^2)\ge 2(x+y+z+t)-1.$

By Jensen's inequality, $4(x^2+y^2+z^2+t^2)\ge (x+y+z+t)^2,\,$ with equality iff $x=y=z=t.\,$ Suffice it to show that

$(x+y+z+t)^2\ge 2(x+y+z+t)-1.$

But this is the same as $(x+y+z+t-1)^2\ge 0,\,$ with the equality for $\displaystyle x=y=z=t=\frac{1}{4}.\,$ To express this in terms of the original variables we solve the system

$\displaystyle \left\{\begin{align}-a+b+c+d=\frac{1}{2}\\a-b+c+d=\frac{1}{2}\\a+b-c+d=\frac{1}{2}\\a+b+c-d=\frac{1}{2}\end{align}\right.$

to obtain $\displaystyle a=b=c=d=\frac{1}{4}.$

### Solution 2

$\displaystyle LHS-RHS=\sum_{cycl}\left(2a-\frac{1}{2}\right)^2\ge 0.$

Equality is attained for $\displaystyle a=b=c=d=\frac{1}{4}.$

### Solution 3

Let $s=a+b+c+d.\,$ Since $f(x)=x^2\,$ is a convex function, Jensen's inequality yields

$\displaystyle \begin{align} \sum_{cycl}(s-2a)^2 &= \sum_{cycl}f(s-2a)\\ &\ge 4f\left(\frac{\displaystyle \sum_{cycl}(s-2a)}{\displaystyle 4}\right)^2\\ &=(a+b+c+d)^2\ge 2(a+b+c+d)-1 \end{align}$

because the latter is equivalent to $(a+b+c+d-1)^2\ge 0.$

Equality for $\displaystyle a=b=c=d=\frac{1}{4}.$

### Solution 4

We have by expanding and regrouping, $\displaystyle \sum_{cycl}(-a+b+c+d)^2=4\sum_{cycl}a^2,\,$ so we need to prove $\displaystyle 4\sum_{cycl}a^2\ge 2\sum_{cycl}a-1.\,$ By the *Power Mean inequality*, $\displaystyle\sqrt{\frac{1}{4}\sum_{cycl}a^2}\ge \frac{1}{4}\sum_{cycl}a.\,$ Thus, suffice it to prove that $\displaystyle \left(\sum_{cycl}a\right)^2\ge 2\left(\sum_{cycl}a\right)-1.\,$

Let $x=a+b+c+d.\,$ We need to prove that $x^2\ge 2x-1,\,$ which is true for all values, with equality at $x=1.\,$ Now we have $a+b+c+d=1\,$ for the minimum which is reached for the convex function for $a=b=c=d=\displaystyle \frac{1}{4}.$

### Solution 5

$\displaystyle \begin{align} LHS - RHS &= 4(a^2+b^2+c^2+d^2)-2(a+b+c+d)+1\\ &=\small{(1^2+1^2+1^2+1^2)(a^2+b^2+c^2+d^2)-2(a+b+c+d)+1}\\ &=(1\cdot a+1\cdot b+1\cdot c+1\cdot d)^2 - 2(a+b+c+d)+1\\ &=(a+b+c+d-1)^2 =0. \end{align}$

Equality: $a=b=c=d\,$ and $a+b+c+d=1,\,$ i.e., $\displaystyle a=b=c=d=\frac{1}{4}.$

### Generalization

We shall prove a more general inequality. Namely,

For integer $n\ge 4,\,$ and $n\,$ real numbers $x_1,\ldots,x_n,$

$\displaystyle \sum_{cycl}(-x_1+x_2+x_3+\ldots+x_n)^2\ge 2\left(\sum_{cycl}x_1\right)^2-1.$

Equaliy occurs if and only if $n=4\,$ and $x_{1}=x_{2}=x_{3}=x_{4}=\frac{1}{4}.\,$ It never occurs for $n\gt 4.$

First of all set $\displaystyle A=\sum_{k=1}^{n}x_{k};\,$ and $\displaystyle B=\sum_{k=1}^{n}x_{k}^{2}.$

and note that by the Cauchy-Schwarz inequality, $\displaystyle B\geq \frac{A^{2}}{n}.$

The inequality can be written as

$(A-2x_{1})^{2}+(A-2x_{2})^{2}+\ldots (A-2x_{n})^{2}\geq 2A-1;$

and we have that

$\begin{align} &(A-2x_{1})^{2}+(A-2x_{2})^{2}+\ldots (A-2x_{n})^{2} \\ &\qquad\qquad=nA^{2}-4A(x_{1}+x_{2}+\ldots +x_{n})+4B \\ &\qquad\qquad=nA^{2}-4A^{2}+4B \\ &\qquad\qquad\geq nA^{2}-4A^{2}+4\frac{A^{2}}{n}. \end{align}$

So suffice it to show that

$\displaystyle nA^{2}-4A^{2}+4\frac{A^{2}}{n}-2A+1\geq 0,$

or

$\begin{align} &n^{2}A^{2}-4nA^{2}+4A^{2}-2An+n \\ &\qquad\qquad =(n-2)^{2}A^{2}-2An+n\geq 0 \end{align}$

Considering the LHS as a second degree polynomial in $A,\,$ we have that the leading coefficient is nonnegative and that

$\displaystyle \Delta =4n^{2}-4n(n-2)^{2}=-4n\left( n-1\right) \left( n-4\right)$

and this is nonpositive for $n\geq 4.\,$ This proves that the polynomial is nonnegative and, in turn, our inequality.

For the equality case note that $\Delta $ must be necessairly $0$ and this happens only if $n=4;$ but it's also necessaty the equality sign in the Cauchy-Schwarz inequality and it occours iff $x_{1}=\ldots =x_{n}$. These two conditions lead to $4A^{2}-8A+4=0\,$ which implies $A=1$ and $x_{1}=\ldots =x_{4}=\frac{1}{4}.$

### Acknowledgment

Dorin Marghidanu has posted this problem at the mathematical inequalities facebook group. Leo Giugiuc has kindly emailed me his solution (Solution 1) with a link. Solution 2 is by Abhay Chandra and, independently, Imad Zak and Ravi Prakash. Mihalcea Andrei Stefan came up with the solution close to #1 where he used the Cauchy-Schwarz inequality instead of Jensen's; Solution 3 is Marian Dinca's variant; Solution 4 is by N.N. Taleb; Solution 5 is by Kunihiko Chikaya. The generalization and its proveare due to Giulio Francot.

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