Matematika wrote: Prove for all positive reals a,b,c,d: $ \frac {a - b}{b + c} + \frac {b - c}{c + d} + \frac {c - d}{d + a} + \frac {d - a}{a + b} \geq 0$ Let $ a\geq b \geq c \geq d$ then $ LHS \geq \frac{a-b}{a+b} +\frac{b-c}{a+b} +\frac{c-d}{a+b} +\frac{d-a}{a+b}=\frac{0}{a+b}=0$

You aren't allowed to assume $ a \geq b\geq c\geq d$ cause inequality isn't symmetric. As much as I know you can't assume that if inequality is cyclic.

Why not , do you have any mathematical explanation for it ?

enndb0x wrote: Matematika wrote: Prove for all positive reals a,b,c,d: $ \frac {a - b}{b + c} + \frac {b - c}{c + d} + \frac {c - d}{d + a} + \frac {d - a}{a + b} \geq 0$ Let $ a\geq b \geq c \geq d$ then $ LHS \geq \frac {a - b}{a + b} + \frac {b - c}{a + b} + \frac {c - d}{a + b} + \frac {d - a}{a + b} = \frac {0}{a + b} = 0$ Can you repeat your argument when I assume $ a\ge c\ge b\ge d$?.My solution : The inequality is equivalent to \[ \frac{a+c}{b+c}+\frac{b+d}{c+d}+\frac{c+a}{d+a}+\frac{d+b}{a+b}\ge 4 \] Apply Cauchy Schwarz inequality in Engel form, we have \[ LHS\ge \frac{(2(a+b+c+d))^2}{(a+c)(b+c)+(b+d)(c+d)+(c+a)(d+a)+(d+b)(a+b)} \] It remains to prove \[ \frac{(2(a+b+c+d))^2}{(a+c)(b+c)+(b+d)(c+d)+(c+a)(d+a)+(d+b)(a+b)} \ge 4 \] But in the fact: we have \[ (a+b+c+d)^2=(a+c)(b+c)+(b+d)(c+d)+(c+a)(d+a)+(d+b)(a+b) \] It ends our proof.

Can you elaborate more how you won that equivalence ?

that was mine solution on competition as well. mathematical proof for that assuming is simply counter example you have expression 2(a-b) (it is cyclic for 2 arguments) and if you could assume $ a \geq b$ you would get that it is bigger then 0 and you can take b less than a and that's not true. and that assumption comes from the fact that in symmetric inequality if you change two variables you don't change a thing but in cyclic ones you do change inequality. just one comment you don't need to add 4 cauchy in engel form is correct for negative values of nominators (denominator must be positive)

There is faster way to prove $ \frac {a + c}{b + c} + \frac {b + d}{c + d} + \frac {c + a}{d + a} + \frac {d + b}{a + b}\ge 4$. We have \[ \frac {a + c}{b + c} + \frac {b + d}{c + d} + \frac {c + a}{d + a} + \frac {d + b}{a + b} = (a + c)(\frac {1}{b + c} + \frac {1}{d + a}) + (b + d)(\frac {1}{c + d} + \frac {1}{a + b})\geq \frac {4(a + b + c + d)}{a + b + c + d} = 4\] which is just $ \frac {1}{a} + \frac {1}{b}\geq \frac {4}{a + b}$ for $ a,b > 0$.

See also here: http://www.artofproblemsolving.com/Forum/viewtopic.php?t=102924

i use cauchy ineequality at first let $ \frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b} =k$ $ k.((a-b)(b+c)+(b-c)(c+d)+(c-d)(d+a)+(d-a)(a+b)) \geq 0$ if a=1 b=2 c=3 d=4 then $ ((a-b)(b+c)+(b-c)(c+d)+(c-d)(d+a)+(d-a)(a+b))<0$then k<0 !!!!![/quote]

table1388 wrote: i use cauchy ineequality at first let $ \frac {a - b}{b + c} + \frac {b - c}{c + d} + \frac {c - d}{d + a} + \frac {d - a}{a + b} = k$ $ k.((a - b)(b + c) + (b - c)(c + d) + (c - d)(d + a) + (d - a)(a + b)) \geq 0$ if a=1 b=2 c=3 d=4 then $ ((a - b)(b + c) + (b - c)(c + d) + (c - d)(d + a) + (d - a)(a + b)) < 0$then k<0 !!!!! [/quote] Cauchy requires positive, or at least non-negative, numbers.

Mathias_DK wrote: table1388 wrote: i use cauchy ineequality at first let $ \frac {a - b}{b + c} + \frac {b - c}{c + d} + \frac {c - d}{d + a} + \frac {d - a}{a + b} = k$ $ k.((a - b)(b + c) + (b - c)(c + d) + (c - d)(d + a) + (d - a)(a + b)) \geq 0$ if a=1 b=2 c=3 d=4 then $ ((a - b)(b + c) + (b - c)(c + d) + (c - d)(d + a) + (d - a)(a + b)) < 0$then k<0 !!!!! Cauchy requires positive, or at least non-negative, numbers.[/quote] i dont think so,i studied my book again arent you wrong ?....

table1388 wrote: Mathias_DK wrote: table1388 wrote: i use cauchy ineequality at first let $ \frac {a - b}{b + c} + \frac {b - c}{c + d} + \frac {c - d}{d + a} + \frac {d - a}{a + b} = k$ $ k.((a - b)(b + c) + (b - c)(c + d) + (c - d)(d + a) + (d - a)(a + b)) \geq 0$ if a=1 b=2 c=3 d=4 then $ ((a - b)(b + c) + (b - c)(c + d) + (c - d)(d + a) + (d - a)(a + b)) < 0$then k<0 !!!!! Cauchy requires positive, or at least non-negative, numbers. i dont think so,i studied my book again arent you wrong ?.... I'm sure You just found a counter example, when they were negative, didn't you?

Mathias_DK wrote: table1388 wrote: Mathias_DK wrote: table1388 wrote: i use cauchy ineequality at first let $ \frac {a - b}{b + c} + \frac {b - c}{c + d} + \frac {c - d}{d + a} + \frac {d - a}{a + b} = k$ $ k.((a - b)(b + c) + (b - c)(c + d) + (c - d)(d + a) + (d - a)(a + b)) \geq 0$ if a=1 b=2 c=3 d=4 then $ ((a - b)(b + c) + (b - c)(c + d) + (c - d)(d + a) + (d - a)(a + b)) < 0$then k<0 !!!!! Cauchy requires positive, or at least non-negative, numbers. i dont think so,i studied my book again arent you wrong ?.... Mathias_DK wrote: I'm sure You just found a counter example, when they were negative, didn't you? yes... .. you are right ...thanks.

I think this problem is from Hung's book and limes123's solution is beautiful enough as given in the book.

Very nice and easy inequality. This is problem equivalently to Baltic Way 1995.

Attachments:
123.doc (33kb)

Zarif wrote: Very nice and easy inequality. This is problem equivalently to Baltic Way 1995. You just reposted a solution...

I`m sorry Pain Rinnegan. I don`t know english and Latex!

I think AG is enough the ineq is equivalent to $\sum\frac{a+c}{b+c}\ge 4$or $(a+b+c+d)(\frac{a+c}{(a+d)(b+c)}+\frac{b+d}{(a+b)(c+d)})\ge 4$ by AG,LHS$\ge 4(a+b+c+d)(\frac{a+c}{(a+d+b+c)^2}+\frac{b+d}{(a+b+c+d)^2})=4$.

We can use trans inequality