Inequalities Help!

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

aa2−bc+1+bb2−ca+1+cc2−ab+1≥1a+b+c\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!

#Algebra

Note by Julian Yu
5 years, 4 months ago

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Comments

We have ,

a2−bc+1=a(a+b+c)+23 [Since ab+bc+ca = 1/3]a^{2} -bc + 1 = a(a + b + c) + \frac{2}{3} \small \text{ [Since ab+bc+ca = 1/3]}

Similarly b2−ca+1=b(a+b+c)+23{ b }^{ 2 }-ca+1 = b(a+b+c) + \frac{2}{3} and c2−ab+1=c(a+b+c)+23c^{2} -ab+1 = c(a+b+c) + \frac{2}{3}

Therefore our inequality becomes 1(a+b+c)+2a3+1(a+b+c)+2b3+1(a+b+c)+2c3≥1a+b+c\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 , 1(a+b+c)+2a3+1(a+b+c)+2b3+1(a+b+c)+2c3≥(1+1+1)25(a+b+c)\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 (1+1+1)25(a+b+c)≥1a+b+c\dfrac{(1+1+1)^{2}}{5(a+b+c)} \ge \dfrac { 1 }{ a+b+c }

since a,b,c are positive reals.

Harsh Shrivastava - 5 years, 4 months ago

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@Julian Yu

Harsh Shrivastava - 5 years, 4 months ago

@Harsh Shrivastava Thanks, but the numerators are a,b,c not 1,1,1.

Julian Yu - 5 years, 4 months ago

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I divided both numerator and denominator by a,b,c.

Harsh Shrivastava - 5 years, 4 months ago

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@Harsh Shrivastava – Oh okay thanks!

Julian Yu - 5 years, 4 months ago

the denominator at the end of the titu's lemma calculation is wrong it should be 11/3 (a+b+c) @Julian Yu @Harsh Shrivastava

T P - 5 years, 4 months ago
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