# 6th European Mathematical Cup (2017), Junior Problem 4

### Problem

### Solution

Observe that, according to Viète's formulas, the polynomial

$\begin{align} f(x)&= x^3 - (y^2 + yz + z^2)x + y^2z+yz^2\\ &=x^3 - [yz-y(y+z)-z(y+z)]x + yz(y + z) \end{align}$

has three real roots: $y,z,-(x+y).$ It follows that $f(x)=(x-y)(x-z)(x+y+z),$ making the required inequality equivalent to

$\displaystyle (x-y)^2(x-z)^2(x+y+z)^2 \le (x^2+y^2+z^2)^3.$

By the AM-GM inequality,

$\displaystyle\begin{align}(x-y)^2(x-z)^2(x+y+z)^2&\le \left(\frac{(x-y)^2+(x-z)^2+(x+y+z)^2}{3}\right)^3\\ &=\left(\frac{3x^2+2y^2+2z^2+2yz}{3}\right)^3, \end{align}$

with equality iff $(x-y)^2=(x-z)^2=(x+y+z)^2.$ Thus suffice it to prove that

$3x^2+2y^2+2z^2+2yz\le3x^2+3y^2+3z^2,$

which is just $(y-z)^2\ge 0,$ with equality for $y=z.$ All the equality conditions hold for $y=z=0$ and $x=\sqrt{3},$ or $y+2x=0,$ i.e., $\displaystyle x=-\frac{1}{\sqrt{3}},$ $\displaystyle y=z=\frac{2}{\sqrt{3}}.$

### Acknowledgment

This is Problem 4 (by Miroslav Marinov) from the 2017 European Mathematical Cup (Junior category). I am grateful to Leo Giugiuc for bringing this competition and the problem to my attention and for supplying the beautiful solution above.

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- An Inequality with Constraint
- An Inequality with Constraints II
- An Inequality with Constraint III
- An Inequality with Constraint IV
- An Inequality with Constraint VII
- An Inequality with Constraint VIII
- An Inequality with Constraint IX
- An Inequality with Constraint X
- Problem 11804 from the AMM
- Sladjan Stankovik's Inequality With Constraint
- An Inequality with Constraint XII
- An Inequality with Constraint XIV
- An Inequality with Constraint XVII
- An Inequality with Constraint in Four Variables II
- An Inequality with Constraint in Four Variables III
- An Inequality with Constraint in Four Variables V
- An Inequality with Constraint in Four Variables VI
- A Cyclic Inequality in Three Variables with Constraint
- Dorin Marghidanu's Cyclic Inequality with Constraint
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints II
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints III
- Inequality with Constraint from Dan Sitaru's Math Phenomenon
- Another Problem from the 2016 Danubius Contest
- Gireaux's Theorem
- An Inequality with a Parameter and a Constraint
- Unsolved Problem from Crux Solved
- An Inequality With Six Variables and Constraints
- Cubes Constrained
- Dorin Marghidanu's Inequality with Constraint
- Dan Sitaru's Integral Inequality with Powers of a Function
- Michael Rozenberg's Inequality in Three Variables with Constraints
- Dan Sitaru's Cyclic Inequality In Three Variables with Constraints IV
- An Inequality with Arbitrary Roots
- Leo Giugiuc's Inequality with Constraint
- Problem From the 2016 IMO Shortlist
- Dan Sitaru's Cyclic Inequality with a Constraint and Cube Roots
- Dan Sitaru's Cyclic Inequality with a Constraint and Cube Roots II
- A Simplified Version of Leo Giugiuc's Inequality from the AMM
- Kunihiko Chikaya's Inequality $\displaystyle \small{\left(\frac{(a^{10}-b^{10})(b^{10}-c^{10})(c^{10}-a^{10})}{(a^{9}+b^{9})(b^{9}+c^{9})(c^{9}+a^{9})}\ge\frac{125}{3}[(a-b)^3+(b-c)^3+(c-a)^3]\right)}$
- A Cyclic Inequality on [-1,1] $\left(xy+yz+zx\ge 1\right)$
- An Inequality with Two Triples of Variables $\left(\displaystyle\sum_{cycl}ux\ge\sqrt{\left(\sum_{cycl}xy\right)\left(2\sum_{cycl}uv-\sum_{cycl}u^2\right)}\right)$
- 6th European Mathematical Cup (2017), Junior Problem 4 $\left(x^3 - (y^2 + yz + z^2)x + y^2z + yz^2 \le 3\sqrt{3}\right)$
- Dorin Marghidanu's Example $\displaystyle\frac{\displaystyle\frac{1}{b_1}+\frac{2}{b_2}+\frac{3}{b_3}}{1+2+3}\ge\frac{1+2+3}{b_1+2b_2+3b_3},$

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