Let a,b,c,d>0. Prove sum (a+2b)/(a+2c) > 4

[NguyenDungTN]

Let a,b,c,d>0. Prove that:
[tex]\sum \frac{a+2b}{a+2c} \ge 4[/tex]

[ins-en]

Please, write your inequality in its full form.

[NguyenDungTN]

I'm sorry!
Prove that:
[tex]\frac{a+2b}{a+2c}+\frac{b+2c}{b+2d}+\frac{c+2d}{c+2a}+\frac{d+2a}{d+2b} \ge 4[/tex]

[dduclam]

I'm sorry!
Prove that:
[tex]\frac{a+2b}{a+2c}+\frac{b+2c}{b+2d}+\frac{c+2d}{c+2a}+\frac{d+2a}{d+2b} \ge 4[/tex]

Using Cauchy-Schwarz inequality

@hello,Dung

[Math Tutor]

Could you write the solution, please?

[dduclam]

Could you write the solution, please?

Ok,here my solution:

[tex]\frac{a+2b}{a+2c}+\frac{b+2c}{b+2d}+\frac{c+2d}{c+2a}+\frac{d+2a}{d+2b}=\sum\frac{a+2b}{a+2c}=\sum\frac{(a+2b)^2}{(a+2b)(a+2c)}[/tex]

[tex]\ge \frac{(a+2b+b+2c+c+2d+d+2a)^2}{\sum(a+2b)(a+2c)}=\frac{9(a+b+c+d)^2}{a^2+b^2+c^2+d^2+6(ab+bc+cd+da)+4(ac+bd)}[/tex]

We only prove [tex]9(a+b+c+d)^2\ge 4(a^2+b^2+c^2+d^2)+24(ab+bc+cd+da)+16(ac+bd)[/tex]which also true by AM-GM

[poster123]

Could you write the solution, please?