I have to prove the following inequality using the Cauchy-Schwarz inequality:
$$\frac{a}{b+c}+\frac{b}{c+d}+\frac{c}{d+a}+\frac{d}{a+b}\ge2$$
where a, b, c and d are positive real numbers.
But I am not able to do it, I am hitting dead-ends with every method I try. Please help!
Answer
By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{a}{b+c}=\sum_{cyc}\frac{a^2}{ab+ac}\geq\frac{(a+b+c+d)^2}{\sum\limits_{cyc}(ab+ac)}=2+\frac{(a+b+c+d)^2-2\sum\limits_{cyc}(ab+ac)}{\sum\limits_{cyc}(ab+ac)}=$$
$$=2+\frac{a^2+c^2+b^2+d^2-2ac-2bd}{\sum\limits_{cyc}(ab+ac)}\geq2+\frac{2\sqrt{a^2c^2}+2\sqrt{b^2d^2}-2ac-2bd}{\sum\limits_{cyc}(ab+ac)}=2.$$
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