# Dorin Marghidanu's Cyclic Inequality with Constraint

### Solution 1

The inequality is equivalent to

$\displaystyle 2a^2-2\sqrt{2}(b+c)a+3b^2+4c^2-2\sqrt{bc}\gt 0.$

But

\begin{align} &2a^2-2\sqrt{2}(b+c)a+3b^2+4c^2-2\sqrt{bc}\\ &\qquad\qquad=(b+c-\sqrt{2}a)^2+(2b^2-2(1+\sqrt{2})bc+3c^2)\\ &\qquad\qquad\ge 2b^2-2(1+\sqrt{2})bc+3c^2. \end{align}

Now, the discriminant of the quadratic form $2b^2-2(1+\sqrt{2})bc+3c^2\,$ equals $2\sqrt{2}-3\lt 0\,$ so that the form is never negative. In addition, it only vanishes for $b=c=0\,$ which is incompatible with the constraint $ab+bc+ca=1.\,$ Therefore, under the constraint, $2b^2-2(1+\sqrt{2})bc+3c^2\gt 0\,$ and the same holds for the whole expression:

$\displaystyle 2a^2+3b^2+4c^2=2a^2-2\sqrt{2}(b+c)a+3b^2+4c^2-2\sqrt{bc}\gt 0.$

### Solution 2

\begin{align} &2a^2+3b^2+4c^2 = (a^2+2b^2)+(b^2+2c^2)+(a^2+2c^2)\\ &\qquad=(a-\sqrt{2}b)^2+(b-\sqrt{2}c)^2+(a-\sqrt{2}c)^2+2\sqrt{2}(ab+bc+ca)\\ &\qquad\ge 2\sqrt{2}\cdot 1=2\sqrt{2}. \end{align}

### Solution 3

By the AM-GM inequality,

$a^2+2b^2\ge 2\sqrt{2}ab,\\ b^2+2c^2\ge 2\sqrt{2}bc.\\ a^2+2c^2\ge 2\sqrt{2}ca.$

Adding up, and taking into account the constraint, yields the required inequality.

### Solution 4

Observe that, for any $\lambda\in (0,3]\,$ and any $\mu\in (0,4],$

$a^2+\lambda b^2\ge 2\sqrt{\lambda}ab,\\ (3-\lambda)b^2+\mu c^2\ge 2\sqrt{(3-\lambda)\mu}bc.\\ a^2+(4-\mu)c^2\ge 2\sqrt{4-\mu}ca.$

By adding the three we have

$2a^2+3b^2+4c^2\ge 2(\sqrt{\lambda}ab+\sqrt{(3-\lambda)\mu}bc+\sqrt{4-\mu}ca).$

To make use of the constraint, we now require $\lambda=(3-\lambda)\mu=4-\mu\,$ which is satisfied with $\lambda=\mu=2.\,$ Thus the three inequalities above become

$a^2+2b^2\ge 2\sqrt{2}ab,\\ b^2+2c^2\ge 2\sqrt{2}bc.\\ a^2+2c^2\ge 2\sqrt{2}ca.$

The first with equality if $a=\pm\sqrt{2}b,\,$ the second becomes equality if $b=\pm\sqrt{2},\,$ and the third, if $a=\pm\sqrt{2}c.\,$ The three are only satified with $a=b=c=0,\,$ which is incompatible with the given constraint. As a consequence, we arrive at the strict inequality

$\displaystyle 2a^2+3b^2+4c^2 \ge 2\sqrt{2}(ab+bc+ca)=2\sqrt{2}.$

### Solution 5

Let $f=2a^2+3b^2+4c^2.\,$ Replacing $\displaystyle a=\frac{1-bc}{b+c},\,$ $f=3b^2+4c^2+\displaystyle\frac{2(1-bc)}{(b+c)^2}.\,$ The derivatives:

$\displaystyle \nabla=\left(\begin{array}{c}\,6b+\frac{4c(-1+bc)}{(b+c)^2}-\frac{4(-1+bc)^2}{(b+c)^3}\\ \frac{4(-1+2c(b+c)^3+b(c+b(-1+bc)))}{(b+c)^3}\end{array}\right).$

Setting $\nabla=0,\,$ we have a tighter bound (going through complex expressions of a real solution) of

$\displaystyle \min(f)=-3+3^{\frac{1}{3}}\left((7-4i\sqrt{2})^{\frac{1}{3}}+(7+4i\sqrt{2})^{\frac{1}{3}}\right)\approx 2.84667,$

which is tighter than $2\sqrt{2}\approx 2.82843.\,$ (The corresponding $b,c\,$ are real.)

wolframalpha gives that value as $\displaystyle 6\cdot \cos\left(\frac{2}{3}\arctan\left(\frac{1}{2\sqrt{2}}\right)\right)-3.$

### Solution 6

\displaystyle\begin{align} &\bigg\{\sqrt{2}a-(b+c)\bigg\}^2+\left\{\sqrt{2}b-\left(1+\frac{1}{\sqrt{2}}\right)c\right\}^2\\ &\qquad&\qquad\qquad +\left\{\left(1-\frac{1}{\sqrt{2}}\right)c\right\}^2+2\sqrt{2}(ab+bc+ca)\\ &\qquad\qquad\ge 2\sqrt{2}(ab+bc+ca)=2\sqrt{2}. \end{align}

Equality could be possible when simultaneously $\sqrt{2}a=b+c,\,$ $\displaystyle \sqrt{bb}=\left(1+\frac{1}{\sqrt{2}}\right)c\,$ $\displaystyle\left(1-\frac{1}{\sqrt{2}}\right)c=0,\,$ $ab+bc+ca=1,\,$ i.e., never as the four conditions are incompatible.

### Solution 7

This solution has been placed on a separate page.

### Acknowledgment

The problem has been kindly posted by Dorin Marghidanu at the CutTheKnotMath facebook page, along with Leo Giugiuc's solution (Solution 1). Originally, the problem appeared at the The School Olympiad facebook group where I found a solution (Solution 2) by Ravi Prakash and another (Solution 3) by Imad Zak; Solution 4 is by Dorin Marghidanu. Athina Kalampoka has independently arrived at Solution 2. The computer assisted Solution 5 is by Nassim Nicholas Taleb; Solution 6 is by Kunihiko Chikaya; Solution 7 is by Le Khanh Sy.