Dorin Marghidanu's Example


Marghidanu's example for application of Bergstrom's inequality

Solution 1

We apply Bergström's Inequality:

$\displaystyle\begin{align} \sum_{k=1}^3\frac{k}{a_k+k} &= \sum_{k=1}^3\frac{k^2}{k(a_k+k)}\\ &\ge\frac{\displaystyle\left(\sum_{k=1}^3k\right)^2}{\displaystyle\sum_{k=1}^3k(a_k+k)}\\ &=\frac{6^2}{\displaystyle\sum_{k=1}^3ka_k+{\displaystyle\sum_{k=1}^3k^2}}\\ &\ge\frac{36}{4+14}\\ &=2. \end{align}$

The equality is achieved when


and $a_1+2a_2+3a_3=4.\;$ Thus, say, $a_1=a_3+2\;$ and $a_2=a_3+1\;$ and, subsequently, $(a_3+2)+2(a_3+1)+3a_3=4.\;$ Solving we get $a_3=0,\;$ $a_2=1,\;$ and $a_1=2\;$ which is easily verified.

Solution 2

By the Cauchy-Schwarz inequality, or by Titu's Lemma,

$\displaystyle\begin{align} LHS &= \frac{1}{a_1+1}+\frac{1}{a_2+2}+\frac{1}{a_2+2}+\frac{1}{a_3+3}+\frac{1}{a_3+3}+\frac{1}{a_3+3}\\ &\ge\frac{(1+1+1+1+1+1)^2}{(a_1+1)+2(a_2+2)+3(a_3+3)}\\ &=\frac{36}{(a_1+2a_2+3a_3)+14}\\ &\ge\frac{36}{4+14}\\ &=2. \end{align}$

The equality occurs when $a_1=2,\;$ $a_2=1,\;$ $a_3=0.$

Solution 3

Setting $b_k=a_k+k\;$ the constraint becomes

$b_1+2b_2+3b_3\le 18,$



The inequality we want to prove is

$\displaystyle\frac{1}{b_1}+\frac{2}{b_2}+\frac{3}{b_3}\ge 2.$

Using Jensen's Inequality with $\displaystyle f(x)=\frac{1}{x}\;$ for $(x\gt 0)\;$ and weights $1,2,3,\;$ we have




We know that in Jensen's inequality the equality occurs iff $b_1=b_2=b_3=b\;$ which in this case means $b\le 3\;$ but, in addition, it is necessary to have $\displaystyle\frac{1}{b_1+2b_2+3b_3}=\frac{1}{18}.$ So that in fact $b=3,\;$ or, in terms of $a_k,\;$ $a_1=2,\;$ $a_2=1\;$ and $a_3=0.$


Dorin Marghidanu has kindly posted the elegant problem above, with solution (Solution 1) that utilized Bergström's inequality, at the CutTheKnotMath facebook page. He later linked to a solution (Solution 2) by Hung Nguyen Viet and posted a solution (Solution 3) by Giulio Francot.


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