User:Imadzak/sandbox

Imad Zak lemma For the positive reals∶a,b,c,m & n and A=b/(ma+nb)+c/(mb+nc)+a/(mc+na)    we have: 1) A≥3/(m+n)     for  m≥2n 2) A>1/n       for m<2n Proof:A=∑▒1/(m a/b+n)=∑▒1/(mx+n )  by letting x=a/b  ,y=b/c&z=c/a==>xyz=1 ==>p=∑▒x ≥3 & q=∑▒xy≥3 by AM-GM We have  A-3/(m+n)=m/(m+n)  1/(∏▒(mx+n) )  B  Where B=-3m^2 (xyz)+(m^2-2mn) ∑▒xy+(2mn-n^2 ) ∑▒x+3n^2=3n^2-3m^2+mq(m-2n)+np(2m-n)≥3n^2-3m^2+3m^2-6mn+6mn-3n^2=0    since m-2n≥0 and 2m-n>0===>A≥3/(m+n)    for m≥2n Q.E.D. Applications∶m=2 &n=1===>∑▒b/(2a+b)≥1….(*) m=3 &n=1===>∑▒b/(3a+b)≥3/4 m=√11& n=√2==>∑▒b/(a√11+b√2)≥3/(√11+√2) For m<2n we  getA=∑▒b/(ma+nb)>∑▒b/(2na+nb)=1/n ∑▒b/(2a+b)>1/n (1)=1/n  by using (*)   7 the inequality is strict Q.E.D. Applications∶m=1 &n=1===>∑▒b/(a+b)>1 Consequence of Imad Zak lemma We have∶ For the positive reals∶a,b,c,m & n and A=b/(ma+nb)+c/(mb+nc)+a/(mc+na)    we have: A≥3/(m+n)     for  m≥2n Let B=a/(ma+nb)+b/(mb+nc)+c/(mc+na)   .Then B≤3/(m+n)     for  m≥2n In fact: mB+nA=3===>mB=3-nA≤3-3n/(m+n)=3m/(m+n)===> B≤3/(m+n) Q.E.D.  and equality holds for a=b=c Conclusion:1)A=b/(ma+nb)+c/(mb+nc)+a/(mc+na) ≥ 3/(m+n)      for  m≥2n 2)  B=a/(ma+nb)+b/(mb+nc)+c/(mc+na)  ≤ 3/(m+n)      for  m≥2n