Cauchy-Schwarz

[dduclam]

Let [tex]a,b,c>0[/tex] .Prove that [tex]\frac a{b}+\frac b{c}+\frac c{a}\ge \frac3{2}+ \frac a{b+c}+\frac b{c+a}+\frac c{a+b}[/tex]

[MM]

It is equivalent to [tex]\frac{(ac)^{2}}{abc(b+c)}+\frac{(ab)^{2}}{abc(c+a)}+\frac{(bc)^{2}}{abc(a+b)}\ge \frac{3}{2}[/tex]. Now we apply Cauchy-Schwartz in Engle form and we obtain [tex]\frac{(ac)^{2}}{abc(b+c)}+\frac{(ab)^{2}}{abc(c+a)}+\frac{(bc)^{2}}{abc(a+b)}\ge \frac{(ab+ac+bc)^{2}}{2abc(a+b+c)}[/tex]. We have to prove that [tex]\frac{(ab+ac+bc)^{2}}{abc(a+b+c)}\ge 3[/tex] which is equivalent to [tex](ab-bc)^{2}+(ac-bc)^{2}+(ab-ac)^{2}\ge 0[/tex] and we're done .
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[MM]

The following generalization also must be true:
[tex]\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}}\ge \frac{n}{2}+\sum_{i=1}^{n}\frac{a_{i}}{a_{i+1}+a_{i+2}}[/tex] holds true for any positive [tex]a_{1},a_{2}\cdots,a_{n}[/tex](in [tex]a_{n+1}[/tex] I mean [tex]a_{1}[/tex] and in [tex]a_{n+2}[/tex] [tex]a_{2}[/tex]).