Exercise 1743Algebra › InequalitiesOlympiadAdded 3 Nov 2026

Cyclic sum of squares

#cyclic sum#squares#factorization

Statement

Prove that for $a,b,c\ge0$ ,
$\frac{a^2+b^2+c^2}{2}\ge\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\cdot\frac{1}{?}$
Hmm, the expression is dimensionally inconsistent. Let me state a correct version:
$\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{a+b+c}{2}.$

Hint

Use Titu's lemma.

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Solution

Titu's lemma.
$\sum\frac{a^2}{b+c}\ge\frac{(a+b+c)^2}{(b+c)+(c+a)+(a+b)}=\frac{(a+b+c)^2}{2(a+b+c)}=\frac{a+b+c}{2}.$

Conclusion. The inequality holds, with equality iff $a=b=c$ .

Cyclic sum of squares

Statement

Solution

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