c/(a+2b)+d/(b+2c)+a/(c+2d)+b/(d+2a)>=c^2/(ac+2bc)+d^2/(bd+2cd)+a^2/(ca+2ad)+b^2/(bd+2ab)....

easy. let $ S=c/{(a + 2b)} + d/{(b + 2c)} + a/{(c + 2d)} + b/{(d + 2a)}$ with caushy: $ \left(c/{(a + 2b)} + d/{(b + 2c)} + a/{(c + 2d)} + b/{(d + 2a)}\right)(2(ab+bc+cd+da+ac+bd))\geq(a+b+c+d)^2$ with caushy: $ 2ab \leq a^2+b^2$ $ 2bc \leq c^2+b^2$ $ 2cd \leq c^2+d^2$ $ 2da \leq a^2+d^2$ $ 2ac \leq a^2+c^2$ $ 2bd \leq d^2+b^2$ summation: $ 2(ab+bc+cd+da+ac+bd)\leq 3(a^2+b^2+c^2+d^2)$ $ \Leftrightarrow 8(ab+bc+cd+da+ac+bd)\leq3(a+b+c+d)^2$ therefore: $ \frac{3}{4}(a+b+c+d)^2S \geq S(2(ab+bc+cd+da+ac+bd))\geq (a+b+c+d)^2$ then: $ S\geq \frac{4}{3}$

orl wrote: For the positive real numbers $ a,b,c$ prove that $ c/{(a + 2b)} + d/{(b + 2c)} + a/{(c + 2d)} + b/{(d + 2a)} > = 4/3$ by Jensen: $ LHS \ge (a+b+c+d)^2\frac{1}{2(ab+ac+ad+bc+bd+cd)} \ge \frac{4}{3}$

Hi Use titu's lemma.

So $\frac c{a + 2b} + \frac d{b + 2c} + \frac a{c + 2d} + \frac b{d + 2a} = \frac {c^2}{ac + 2bc} + \frac {d^2}{bd + 2cd} + \frac {a^2}{ca + 2da} + \frac {b^2}{db + 2ab} \geq$ $ \dfrac {(\sum a)^2} {2\sum ab} \geq \dfrac {4}{3}$, since $\left ( \dfrac {\sum a}{4} \right )^2 \geq \dfrac {\sum ab}{6}$, by the Maclaurin inequality.

Positive $a,b,c $ or $a,b,c,d $?

it is just Cauchy Schwartz

Sorry. Thanks Skravin.

making the form into $\sum_{cyc} \frac{1}{ \frac{a}{c}+\frac{2b}{c} } $ then use T2+Cauchy

sqing wrote: orl wrote: For the positive real numbers $ a,b,c$ prove that \[ \frac c{a + 2b} + \frac d{b + 2c} + \frac a{c + 2d} + \frac b{d + 2a} \geq \frac 43.\] See also here http://www.artofproblemsolving.com/community/c6h246429p1353592 ??? The link is here

or using the WLOG condition between a,b,c,d then using the rearrangement theorem and Cauchy

Skravin wrote: or using the WLOG condition between a,b,c,d then using the rearrangement theorem and Cauchy For assuming a condition of comparing you must see if a cyclic transformation holds good.

adityaguharoy wrote: Skravin wrote: or using the WLOG condition between a,b,c,d then using the rearrangement theorem and Cauchy For assuming a condition of comparing you must see if a cyclic transformation holds good. oh... I missed it

$3(a^2+b^2+c^2+d^2)>=2(ab+ac+ad+bc+bd+dc)$ Use AM-GM: $a^2+b^2>=2ab$ $a^2+c^2>=2ac$ $a^2+d^2>=2ad$ $b^2+c^2>=2bc$ $b^2+d^2>=2bd$ $c^2+d^2>=2cd$