# Inequality with Three Linear Constraints

### Proof 1

Denote $a+b=s,\;$ $ab=p.\;$ Then $0\lt s\le 5\;$ and $c\le 11-s.\;$ Hence,

$\displaystyle\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{s}{p}+\frac{1}{11-s}.$

Thus, suffice it to show that $\displaystyle\frac{s}{p}+\frac{1}{11-s}\ge 1.$

That inequality is equivalent to $11s-s^2-p(10-s)\ge 0.\;$ We'll maximize $p.$

For any fixed $s\in[0,5],\;$ define, for $a\in (0,\min\{2,s\}],\;$ quadratic function $f_s(a)=a(s-a).\;$ We consider two cases:

$\mathbf{s\le 4}$

By the AM-GM inequality, $a(s-a)\le\displaystyle\frac{s^2}{4},\;$ implying $p\le\displaystyle\frac{s^2}{4}.\;$ It follows that

\displaystyle\begin{align} 11s-s^2-p(10-s) &\ge 11s-s^2-\frac{s^2}{4}(10-s)\\ &=\frac{s[s-(7-\sqrt{5})][s-(7+\sqrt{5})]}{4}\\ &\gt 0. \end{align}

$\mathbf{4\lt s\le 5}$

Function $f_s\;$ is strictly increasing on $(0,2]\;$ so that

$\max p=\max f_s=f_s(2)=2(s-2).$

It follows that

\displaystyle\begin{align} 11s-s^2-p(10-s) &\ge 11s-s^2+2(s-2)(s-10)\\ &=(s-5)(s-8)\\ &\ge 0. \end{align}

### Proof 2

Set $x=3a,\;$ $y=2b,\;$ and $z=c.\;$ Then the constraints become

(1)

$x\le 6,$

(2)

$2x+3y\le 30,$

(3)

$2x+3y+6z\le 66.$

By the AM-GM inequality, (2) implies

$30\ge 2x+3y =x+x+y+y+y\ge 5\sqrt[5]{x^2y^3},$

so that $6\ge\sqrt[5]{x^2y^3},\;$ i.e.,

(4)

$x^2y^3\le 6^5.$

From (3), $66\ge 11\sqrt[11]{x^2y^3z^6},\;$ so that

(5)

$x^2y^3z^6\le 6^11.$

Now, multiplying (1) (twice), (4), and (5) gives $(xyz)^6\le 6^{18},\;$ or

(6)

$xyz\le 6^3.$

Further, multiplying (1) and (4) gives $(xy)^3\le 6^6,\;$ or

(7)

$xy\le 6^2.$

Now, multiply (1), (6), and (7) to get

(8)

$x^3y^2z\le 6^6.$

Finally, using AM-GM inequality once more, with a reference to (8),

\displaystyle\begin{align} \frac{1}{a}+\frac{1}{b}+\frac{1}{c} &= \frac{3}{x}+\frac{2}{y}+\frac{1}{z}\\ &\ge 6\sqrt[6]{\frac{1}{x^3y^2z}}\\ &\ge 6\sqrt[6]{\frac{1}{6^6}}\\ &=1. \end{align}

Equality is achieved for $x=y=z,\;$ or $3a=2b=c,\;$ i.e., $(a,b,c)=(2t,3t,6t),\;$ $t\in [0,1].$

### Proof 3

It can be verified directly or with the Taylor expansions that

\displaystyle\begin{align} \frac{1}{a} &= \frac{1}{2}+\frac{2-a}{4}+\frac{(2-a)^2}{4a},\\ \frac{1}{b} &= \frac{1}{3}+\frac{3-b}{9}+\frac{(3-b)^2}{9b},\\ \frac{1}{c} &= \frac{1}{6}+\frac{6-c}{36}+\frac{(6-c)^2}{36c}.\\ \end{align}

Now,

\displaystyle\begin{align} \frac{1}{a}+\frac{1}{b}+\frac{1}{c} &= \left(\frac{1}{2}+\frac{2-a}{4}+\frac{(2-a)^2}{4a}\right)\\ &\qquad+\left(\frac{1}{3}+\frac{3-b}{9}+\frac{(3-b)^2}{9b}\right)+= \frac{1}{6}+\frac{6-c}{36}+\frac{(6-c)^2}{36c}\\ &=1+\frac{5(2-a)+3(5-(a+b))+(11-(a+b+c))}{36}\\ &\qquad+\frac{(2-a)^2}{4a}+\frac{(3-b)^2}{9b}+\frac{(6-c)^2}{36c}\\ &\ge 1, \end{align}

as the last four terms are not-negative.

### Acknowledgment

The problem (by Nguyen Viet Hung) was communicated to me by Leo Giugiuc, along with his solution (Proof 1). Imad Zak has replied with a different solution (Proof 2). Proof 3 is an example of a more general technique found in Martin Celli's article Convexity, and Hung’s inequality with linear constraints from the Romanian Mathematical Magazine, Jan 27, 2017.