# Dorin Marghidanu's Inequality with Constraint

### Solution 1

Set $a_1+1=x\,$ $2a_2+1=y,\,$ $3a_3+1=z.\,$ Then $x+2y+3z\le 9\,$ and we need to prove that $\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}\ge 4.$

$\displaystyle \left(\frac{1}{x}+\frac{2}{y}+\frac{3}{z}\right)(x+2y+3z)\ge (1+2+3)^2=36,$

i.e., $\displaystyle \frac{1}{x}+\frac{2}{y}+\frac{3}{z}\ge\frac{36}{x+2y+3z}\ge\frac{36}{9}=4.$

Equality holds iff $x+2y+3z=9\,$ and, simultaneously, $\frac{\displaystyle x}{\displaystyle \frac{1}{x}}=\frac{\displaystyle 2y}{\displaystyle \frac{2}{y}}=\frac{\displaystyle 3z}{\displaystyle \frac{3}{z}},\,$ which is equivalent to $\displaystyle x=y=z=\frac{3}{2},\,$ so that $\displaystyle a_1=\frac{1}{2},\,$ $\displaystyle a_2=\frac{1}{4},\,$ $\displaystyle a_3=\frac{1}{6}.$

### Solution 2

Using (weighted) Jensen's inequality for the convex function $\displaystyle f(x)=\frac{1}{x},\,$ we get

\displaystyle \begin{align} &\frac{1}{a_1+1}+\frac{2}{2a_2+1}+\frac{3}{3a_3+1}=f(a_1+1)+2f(a_2+1)+3f(3a_3+1)\\ &\qquad\qquad\qquad\ge(1+2+3)f\left(\frac{(a_1+1)+2(2a_2+1)+3(3a_3+1)}{1+2+3}\right)\\ &\qquad\qquad\qquad=\frac{\displaystyle 6}{\displaystyle \frac{a_1+4a_2+9a_3+6}{6}}\\ &\qquad\qquad\qquad=\frac{36}{a_1+4a_2+9a_3+6}\ge\frac{36}{9}=4. \end{align}

Equality is attained for $a_1+1=2a_2+1=3a_3+1,\,$ i.e., for $a_1=2a_2=3a_3.\,$ From $a_1+3a_2+9a_3\le 3,\,$ $6a_1\le 3,\,$ meaning $\displaystyle a_1\le \frac{1}{2}.\,$ On the other hand, $\displaystyle \frac{1}{a_1+1}+\frac{2}{2a_2+1}+\frac{3}{3a_3+1}\ge 4,\,$ implies

$\displaystyle \frac{1}{a_1+1}+\frac{2}{a_1+1}+\frac{3}{a_1+1}=\frac{6}{a_1+1}\ge 4,$

i.e., $\displaystyle a_1\ge\frac{1}{2}.\,$ Hence, for equality, $\displaystyle a_1=\frac{1}{2},\,$ $\displaystyle a_2=\frac{1}{4},\,$ $\displaystyle a_3=\frac{1}{6}.$

### Solution 3

Let $a_1=x-1,\,$ $\displaystyle a_2=\frac{y-1}{2},\,$ $\displaystyle a_3=\frac{z-1}{3}.\,$ We are looking at $\displaystyle \min\left[\frac{1}{x}+\frac{2}{y}+\frac{3}{z}\right]\,$ under the constraints $x\gt 0,\,$ $y\gt 0,\,$ $z\gt 0,\,$ $x+2y+3z\le 9.\,$

The minimum is reached for maximal values of $x,y,z,\,$ hence we trun to $x+2y+3z=9\,$ and use the Lagrange multipliers:

$\displaystyle \nabla=\begin{pmatrix}-\lambda-\frac{1}{x^2}\\-2 \lambda-\frac{2}{y^2}\\-3 \lambda-\frac{3}{z^2}\\9-x-2y-3z\end{pmatrix}=0,$

from which $\displaystyle x=y=z=\frac{3}{2},\,$ with $\displaystyle \min\left[\frac{1}{x}+\frac{2}{y}+\frac{3}{z}\right]=4.\,$ For the original variables, $\displaystyle a_1=\frac{1}{2},\,$ $\displaystyle a_2=\frac{1}{4},\,$ $\displaystyle a_3=\frac{1}{6}.$

### Solution 4

Let $a_1=x-1,\,$ $\displaystyle a_2=\frac{y-1}{2},\,$ $\displaystyle a_3=\frac{z-1}{3}.\,$ We are looking at $\displaystyle \min\left[\frac{1}{x}+\frac{2}{y}+\frac{3}{z}\right]\,$ under the constraints $x\gt 0,\,$ $y\gt 0,\,$ $z\gt 0,\,$ $x+2y+3z\le 9.\,$

The minimum is reached for maximal values of $x,y,z,\,$ hence we trun to $x+2y+3z=9.\,$ The function

$\displaystyle f=\frac{1}{x}+\frac{2}{y}+\frac{3}{z}=\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{z}$

is convex, it has for minimum the smallest dispaerion among the variables, so $x=y=z\,$ at their maximal values, i.e., $\displaystyle x=y=z=\frac{3}{2}.$

### Acknowledgment

This problem has been kindly posted at the CutTheKnotMath facebook page by Dorin Marghidanu. Solution 1 is by Leo Giugiuc; Solution 2 is Marian Dinca; Solutions 3 and 4 are by N. N. Taleb.