To prove the inequality (\frac{a}{b+c} + \frac{a+c} + \frac{c}{a+b} \geq \frac{3}{2}) for positive real numbers (a, b, c), we can use Titu's Lemma (a special case of Cauchy-Schwarz inequality) and basic algebraic manipulations. Here's the step-by-step proof:
Titu's Lemma states:
[ \sum_{i=1}^n \frac{x_i^2}{yi} \geq \frac{(\sum{i=1}^n xi)^2}{\sum{i=1}^n y_i} \quad \text{for positive } x_i, y_i ]
Rewrite the left-hand side (LHS) of the inequality as:
[ \frac{a}{b+c} + \frac{a+c} + \frac{c}{a+b} = \frac{a^2}{a(b+c)} + \frac{b^2}{b(a+c)} + \frac{c^2}{c(a+b)} ]
By Titu's Lemma:
[ \geq \frac{(a + b + c)^2}{a(b+c) + b(a+c) + c(a+b)} ]
The denominator in the RHS is:
[ a(b+c) + b(a+c) + c(a+b) = ab + ac + ab + bc + ac + bc = 2(ab + bc + ca) ]
Thus:
[ \geq \frac{(a + b + c)^2}{2(ab + bc + ca)} ]
We know:
[ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) \geq 3(ab + bc + ca) ]
Substitute this into the expression:
[ \frac{(a + b + c)^2}{2(ab + bc + ca)} \geq \frac{3(ab + bc + ca)}{2(ab + bc + ca)} = \frac{3}{2} ]
The inequality holds:
[ \boxed{\frac{a}{b+c} + \frac{a+c} + \frac{c}{a+b} \geq \frac{3}{2}} ]
Equality occurs when (a = b = c).
Answer: (\boxed{\dfrac{3}{2}}) (since the question asks to prove the inequality, the minimal value is (\frac{3}{2}))
(\boxed{\dfrac{3}{2}})
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