Maths Olympiad Prep

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Algebra Difficulty 6.1 National olympiad Prove it

8. (1) Given that a,b,ca, b, c are positive numbers, prove that ab+c+bc+a+ca+b32\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}. (1963 Moscow Mathematical Olympiad Problem)

Solution

8. (1) It is easy to see that the inequality to be proved is equivalent to 2(a3+b3+c3)a2b+a2c+b2a+b2c+2\left(a^{3}+b^{3}+c^{3}\right) \geqslant a^{2} b+a^{2} c+b^{2} a+b^{2} c+ c2a+c2bc^{2} a+c^{2} b. Since a,b,ca, b, c are positive real numbers, we have
a3+b3a2b+ab2,b3+c3b2c+bc2,c3+a3c2a+ca2a^{3}+b^{3} \geqslant a^{2} b+a b^{2}, b^{3}+c^{3} \geqslant b^{2} c+b c^{2}, c^{3}+a^{3} \geqslant c^{2} a+c a^{2}

Adding the three inequalities yields the desired result.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.