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The Problems of Ukrainian Authors

Ukraine algebra

Problem

For positive numbers

a

,

b

with

a + b = ab

prove inequality:

\frac{a}{b ^{2} + 4} + \frac{b}{a ^{2} + 4} \geq \frac{1}{2} .

Solution

ab = a + b \geq 2 ab \Rightarrow ab \geq 4

, then

\frac{a}{b ^{2} + 4} + \frac{b}{a ^{2} + 4} \geq \frac{a}{b ^{2} + ab} + \frac{b}{a ^{2} + ab} = = \frac{a}{b ( a + b )} + \frac{b}{a ( a + b )} = \frac{a ^{2} + b ^{2}}{( a + b ) ^{2}} \geq \frac{1}{2} .

Techniques

QM-AM-GM-HM / Power MeanCauchy-Schwarz