Given \(a, b, c\) are positive real numbers such that \(ab+bc+ca=\frac { 1 }{ 3 } \), prove the following inequality:

\[\large \frac { a }{ { a }^{ 2 }-bc+1 } +\frac { b }{ { b }^{ 2 }-ca+1 } +\frac { c }{ { c }^{ 2 }-ab+1 } \ge \frac { 1 }{ a+b+c } \]

Any help would be appreciated, thanks!

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TopNewestWe have ,

\(a^{2} -bc + 1 = a(a + b + c) + \frac{2}{3} \small \text{ [Since ab+bc+ca = 1/3]}\)

Similarly \({ b }^{ 2 }-ca+1 = b(a+b+c) + \frac{2}{3}\) and \(c^{2} -ab+1 = c(a+b+c) + \frac{2}{3}\)

Therefore our inequality becomes \(\large \frac { 1}{ (a + b + c) + \frac{2a}{3} } +\frac { 1 }{ (a + b + c) + \frac{2b}{3}} +\frac { 1 }{ (a + b + c) + \frac{2c}{3}} \ge \frac { 1 }{ a+b+c } \)

Now applying Titu's Lemma , \[\large \frac { 1}{ (a + b + c) + \frac{2a}{3} } +\frac { 1 }{ (a + b + c) + \frac{2b}{3}} +\frac { 1 }{ (a + b + c) + \frac{2c}{3}} \ge \dfrac{(1+1+1)^{2}}{5(a+b+c)} \]

Now it is obvious that \(\dfrac{(1+1+1)^{2}}{5(a+b+c)} \ge \dfrac { 1 }{ a+b+c } \)

since a,b,c are positive reals. – Harsh Shrivastava · 1 year, 5 months ago

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@Julian Yu @Harsh Shrivastava – T P · 1 year, 4 months ago

the denominator at the end of the titu's lemma calculation is wrong it should be 11/3 (a+b+c)Log in to reply

@Harsh Shrivastava Thanks, but the numerators are a,b,c not 1,1,1. – Julian Yu · 1 year, 5 months ago

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– Harsh Shrivastava · 1 year, 5 months ago

I divided both numerator and denominator by a,b,c.Log in to reply

– Julian Yu · 1 year, 5 months ago

Oh okay thanks!Log in to reply

@Julian Yu – Harsh Shrivastava · 1 year, 5 months ago

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