MathDB
a+b+c = 3

Source: China North MO

August 14, 2006
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Problem Statement

a,b,ca,b,c are positive numbers such that a+b+c=3a+b+c=3, show that: a2+92a2+(b+c)2+b2+92b2+(a+c)2+c2+92c2+(a+b)25\frac{a^{2}+9}{2a^{2}+(b+c)^{2}}+\frac{b^{2}+9}{2b^{2}+(a+c)^{2}}+\frac{c^{2}+9}{2c^{2}+(a+b)^{2}}\leq 5