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Baltic Way 2021 · Problem 2

Algebra

Let a,b,ca, b, c be the side lengths of a triangle. Prove that

(a2+bc)(b2+ca)(c2+ab)3>a2+b2+c22.\sqrt[3]{\left(a^{2}+b c\right)\left(b^{2}+c a\right)\left(c^{2}+a b\right)}>\frac{a^{2}+b^{2}+c^{2}}{2} .
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Topics

Equations and inequalities

Solutions

Solution

We claim that

a2+bc>a2+b2+c22,a^2 + bc > \frac{a^2 + b^2 + c^2}{2},

which will finish the proof. Note that the claimed inequality is equivalent to

a2+bc>a2+b2+c22  ⟺  2a2+2bc>a2+b2+c2  ⟺  a2>(b−c)2  ⟺  a>∣b−c∣,\begin{aligned} a^2 + bc > \frac{a^2 + b^2 + c^2}{2} &\iff 2a^2 + 2bc > a^2 + b^2 + c^2 \\ &\iff a^2 > (b-c)^2 \iff a > |b-c|, \end{aligned}

which holds due to the assumption of a,b,ca, b, c being side lengths of a triangle.

Contest context

Results from Baltic Way 2021

12 teams

Mean score
3.3 / 5
Scores of 4 or 5
8 / 12
Estonia
5 / 5

Score distribution

04
10
20
30
41
57
All team scores
TeamScore
St. Petersburg5 / 5
Estonia5 / 5
Germany5 / 5
Latvia5 / 5
Lithuania5 / 5
Poland0 / 5
Denmark0 / 5
Norway5 / 5
Finland4 / 5
Sweden0 / 5
Iceland0 / 5
Ireland5 / 5