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Algebra Difficulty 5.9 AIME, harder Prove it

Example 5 Proof: For positive real numbers a,b,ca, b, c, we have ab+c+bc+a+ca+b⩾32\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \geqslant \frac{3}{2}.

Solutions — 2

Solution 1

Prove that using the Cauchy-Schwarz inequality, we have
((b+c)+(c+a)+(a+b))(1b+c+1c+a+1a+b)⩾32,((b+c)+(c+a)+(a+b))\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \geqslant 3^{2},

so
a+b+cb+c+a+b+cc+a+a+b+ca+b⩾92\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b} \geqslant \frac{9}{2}

thus inequality (8) holds.

Solution 2

Prove that using the Cauchy-Schwarz inequality, we have
((b+c)+(c+a)+(a+b))(1b+c+1c+a+1a+b)⩾32((b+c)+(c+a)+(a+b))\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right) \geqslant 3^{2}

Therefore,
a+b+cb+c+a+b+cc+a+a+b+ca+b⩾92\frac{a+b+c}{b+c}+\frac{a+b+c}{c+a}+\frac{a+b+c}{a+b} \geqslant \frac{9}{2}

Thus, inequality (8) holds.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.