# An Inequality With Six Variables and Constraints

### Problem ### Solution 1

Let $x=a-1,\,y=b-1,\,z=c-1.\,$ Then $x+y+z+d+e+f=1,\,$ where all variables are not negative. By the Cauchy-Schwarz inequality,

\displaystyle \begin{align} &(x+1)^2+(y+1)^2+(z+1)^2+d^2+e^2+f^2\\ &\qquad\qquad\qquad\ge\frac{\left(3[(x+1)+(y+1)+(z+1)]+(d+e+f)\right)^2)}{30}\\ &\qquad\qquad\qquad=\frac{(2(x+y+z)+10)^2}{30}. \end{align}

The RHS has a minimum when $x+y+z=0\,$ equal to $\displaystyle \frac{10^2}{30}=\frac{10}{3}.\,$ It is achieved for $x=y=z=0\,$ and $\displaystyle d=e=f=\frac{1}{3}.$

On the other hand,

\displaystyle \begin{align} &(x+1)^2+(y+1)^2+(z+1)^2+d^2+e^2+f^2\\ &\qquad\qquad\qquad =x^2+y^2+z^2+d^2+e^2+f^2+2(x+y+z)+3. \end{align}

Since $d,e,f\ge 0,$

\begin{align} &d^2+e^2+f^2\le (1-x-y-z)^2\\ &(x+1)^2+(y+1)^2+(z+1)^2+d^2+e^2+f^2\\ &\qquad\qquad\qquad\le 4+2(x^2+y^2+z^2). \end{align}

Equality occurs when $d=e=f=0,\,$ so we need to minimize the RHS for $x+y+z=1:$

\begin{align} x^2+y^2+z^2&= (1-y-z)^2+y^2+z^2\\ x^2+y^2+z^2&=1+2(y^2+z^2-y-z+yz)\\ &=1+2[-y(1-y-z)-z(1-z)]\\ &\le 1. \end{align}

We get that

$(x+1)^2+(y+1)^2+(z+1)^2+d^2+e^2+f^2\le 6,$

with equality for $x=1,\,$ $y=z=d=e=f=0.$

### Solution 2

Let $\displaystyle P=\sum_{all}a^2.\,$ By the Cauchy-Schwarz inequality,

$\displaystyle a^2+b^2+c^2\ge\frac{1}{3}(a+b+c)^2=\frac{t^2}{3},\;\text{with}\,3\le t\le 4,\\ \displaystyle d^2+e^2+f^2\ge\frac{1}{3}(d+e+f)^2=\frac{(4-t)^2}{3}\\ \displaystyle P\ge\frac{t^2+(4-t)^2}{3}=\frac{2(t-2)^2+8}{3}\ge\frac{10}{3},\;\text{for}\, 3\le t\le 4.$

It follows that $\min P\displaystyle =\frac{10}{3},\,$ where $a=b=c=1,\,$ $\displaystyle d=e=f=\frac{1}{3}.$

Further,

(*)

\begin{align} P&=(a+b+c)^2-2(ab+bc+ca)\\ &\qquad\qquad\qquad+(d+e+f)^2-2(de+ef+fd)\\ &\le t^2+(4-t)^2-2(ab+bc+ca). \end{align}

Since $a,b,c\ge 1,\,$ we also have

$(a-1)(b-1)+(b-1)(c-1)+(c-1)(a-1)\ge 0,$

implying,

(**)

$ab+bc+ca\ge 2(a+b+c)-3.$

From (*) and (**) it follows that

\begin{align} P&\le (4-t)^2+t^2-4t+6=2t^2-12t+22\\ &=2(t-3)^2+4\le 6,\;\text{for}\, 3\le t\le 4. \end{align}

Thus, $\max P=6,\,$ where $t=4,\,$ i.e., $a=2,\,$ $b=c=1,\,$ $d=e=f=0.$

### Solution 3

Convexity argument: let $F=a^2+b^2+c^2+d^2+e^2+f^2.\,$ We can show that $\displaystyle \frac{10}{3}\le F\le 6.\,$ Since the square function is convex and $x^2\le x\,$ for $0\le x\le 1,\,$ and $x^2\ge x\,$ for $x\ge 1,\,$ we get the maximum $F=6\,$ for the maximum dispersion under the constraints: $\{2,1,1,0,0,0\}\,$ (or maximum variance) and we get the minimum $F\displaystyle =\frac{10}{3}\,$ under the minimum variance under the constraint: $\{1,1,1,\displaystyle \frac{1}{3},\frac{1}{3},\frac{1}{3}\}.$

### Generalization

Let $n\,$ be a natural number, $n\ge 2,\,$ $\{a_1,a_2,\ldots,a_n\}\subset [0,1]\,$ and $\{b_1,b_2,\ldots,b_n\}\subset [1,\infty)\,$ such

$\displaystyle \sum_{k=1}^na_n+\sum_{k=1}^nb_n=n+1.$

Prove that

$\displaystyle \frac{n^2+1}{n}\le\sum_{k=1}^na_n^2+\sum_{k=1}^nb_n^2\le n+3.$

### Solution 4

By the Cauchy-Scharz inequality,

$\displaystyle \sum_{k=1}^na_n^2+\sum_{k=1}^nb_n^2\ge\frac\left(\sum_{k=1}^na_k+n\sum_{k=1}^nb^k\right)^2}{n+n^3}$

The substitution $\displaystyle \displaystyle \sum_{k=1}^na_k=n+1-\sum_{k=1}^nb_k,\,$ we get

$\displaystyle \sum_{k=1}^na_n^2+\sum_{k=1}^nb_n^2\ge\frac\left(n+1+(n-1)\sum_{k=1}^nb^k\right)^2}{n+n^3}$

Minimum of the RHS is obtained when $\displaystyle \sum_{k=1}^nb_k=n:$

\displaystyle\begin{align} \sum_{k=1}^na_k^2+\sum_{k=1}^nb_k^2 &\ge\frac\left(n+1+(n-1)n\right)^2}{n+n^3}\\ &=\frac{(n^2+1)^2}{n(n^2+1)}=\frac{n^2+1}{n}. \end{align

Therefore we get the LHS inequality. The equality occurs when $\displaystyle a_k=\frac{1}{k}\,$ and $b_k=1.$

For the right inequality, observe that, since $a_k\ge 0,$

\displaystyle\begin{align} \sum_{k=1}^na_k^2&\le\left(\sum_{k=1}^na_k\right)^2=\left(n+1-\sum_{k=1}^nb_k\right)^2\\ &=\left(1-\sum_{k=1}^n(b_k-1)\right)^2. \end{align}

Therefore,

$\displaystyle\sum_{k=1}^na_k^2+\sum_{k=1}^nb_k^2\le \left(1-\sum_{k=1}^n(b_k-1)\right)2+\sum_{k=1}^n(b_k-1+1)^2.$

We need to maximize the RHS with $\displaystyle \sum_{k=1}^n(b_k-1)\le 1.\,$ Let $b_k-1=x_k,\,$ then $\displaystyle \sum_{k=1}^nx_k\le 1\,$ and the RHS can be simplified as follows

\displaystyle\begin{align} &\left(1-\sum_{k=1}^nx_k\right)^2+\sum_{k=1}^n(x_k+1)\\ &\qquad\qquad\qquad=\left(\sum_{k=1}^nx_k\right)^2+\sum_{k=1}^nx_k^2+n+1\\ &\qquad\qquad\qquad\le n+2+\sum_{k=1}^nx_k^2. \end{align}

We have, $\displaystyle \sum_{k=1}^nx_k^2\le\left(\sum_{k=1}^nx_k\right)^2=1,\,$ with equality at $x_1=1\,$ $x_k=0,\,$ $k\ne 1.\,$ Therefore,

$\displaystyle \sum_{k=1}^na_k^2+\sum_{k=1}^nb_k^2\le n+3,$

with equality at $b_1=2,b_k=0,k\ne 1\,$ and $a_k=0,\,$ for all $k.$

### Solution 5

Since $\forall a_i\in [0,1],\,$ $a_i^2\le a_i,\,$ implying

\displaystyle \begin{align} \sum_{k=1}^na_k^2+\sum_{k=1}^nb_k^2 &\le \sum_{k=1}^na_k+\sum_{k=1}^nb_n^2\\ &=\sum_{k=1}^nb_k^2+n+1-\sum_{k=1}^nb_k\\ &=\sum_{k=1}^n[(b_k^2-1)^2+(b_k-1)]+n+1\\ \end{align}

We would like to show that $\displaystyle \sum_{k=1}^n[(b_k^2-1)^2+(b_k-1)]+n+1\le n+3,\,$ i.e., that

(1)

$\displaystyle \sum_{k=1}^n[(b_k^2-1)^2+(b_k-1)]\le 2.$

Let's see. $b_k\ge 1,\,$ hence $b_k^2\ge b_k\,$ and $\displaystyle \sum_{k=1}^nb_k\ge n,\,$ so that $\displaystyle\sum_{k=1}^n(b_k-1)\ge 0.$

On the other hand, $\displaystyle \sum_{k=1}^na_k+\sum_{k=1}^nb_k=n+1,\,$ so that $\displaystyle \sum_{k=1}^nb_k\le n+1\,$ and subsequently $\displaystyle \sum_{k=1}^n(b_k-1)\le 1.$

\displaystyle\begin{align} &b_k-1\le 1\,\Rightarrow\, (b_k-1)^2\le b_k-1\,\Rightarrow\\ &\sum_{k=1}^n(b_k-1)^2\le\sum_{k=1}^n(b_k-1))\,\Rightarrow\\ &\sum_{k=1}^n[(b_k-1)^2+(b_k-1)]\le 2 \end{align}

implying (1).

### Acknowledgment

I am grateful to Leo Giugiuc who has kindly posted the problem of his at the CutTheKnotMath facebook page and encouraged Siva Siva Modugula to comment (Solution 1). Leo also commented with a link to a solution by Richdad Phuc (Solution 2) at the mathematical inequalities facebook group and pointed to generalizzations. Solution 3 is by N. N. Taleb.

The generalization is actually Leo's original problem. Solution 4 (only RHS) is by Siva Modugula; Solution 5 (only LHS) is by Bunget Mihai. 