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Saturday, February 27, 2021

2021/011) Given positive a,b,c show that a^2(b+c) + b^2(a+c) + c^2(a+b) >= 6abc

 We have 

LHS = abc(\frac{a}{c} + \frac{a}{b}) +  abc(\frac{b}{c} + \frac{b}{a})+ abc(\frac{c}{b} + \frac{c}{a})

= abc(\frac{a}{c} + \frac{a}{b} + \frac{b}{c} + \frac{b}{a} + \frac{c}{b} + \frac{c}{a})

= abc(\frac{a}{c} + \frac{c}{a} + \frac{b}{c} + \frac{c}{b} + \frac{a}{b} + \frac{b}{a})

as \frac{a}{c} + \frac{c}{a} \ge 2  

\frac{b}{c} + \frac{c}{b} \ge 2

\frac{a}{b} + \frac{b}{a} \ge 2

The above 3 relations follow from am gm inequality


putting the value we get

LHS >= abc(2+2+2) or LHS >= 6abc 


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