1] Let a,b,c be non-negative real numbers such that a+b+c=3. Prove that
\frac a{a+2bc}+\frac b{b+2ca}+\frac c{c+2ab}\ge1
341@dimensions - what is your argument after you prove that
LHS ≤ 3/(2+bc/a)?
It is not possible to establish that 3/(2+bc/a) ≤1 as that would mean that a≤bc which is not consistent with a≥b≥bc (as c≤1 which follows from the assumption that a≥b≥c and a+b+c=3)
39i got it at last using cauchy schwarz (after almost 2 hours!)
by cauchy schwarz
\[ \sum_{cyclic}\frac{bc}{2a+bc}=\sum_{cyclic}\frac{(bc)^{2}}{2abc+b^{2}c^{2}}\geq\frac{(ab+bc+ca)^{2}}{6abc+a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}} \]
\[ =\frac{a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}+2abc(a+b+c)}{6abc+a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}}=1 \]
39can u suggest any book or website that could help me out with jensen inequality?
i mean i wanna become familiar with it like cauchy shcwarz
39i know jensen but never get the right method to use it like u ...........
i have been trying for cachy shcwarz since long time , but havent succeded.
anyways perfect soln sir
341First I tried using Cauchy Schwarz but found that the inequality was tighter than that.
Jensen on the function 1/1+2x2 seemed to be what was needed, but this function is not convex on the entire interval (0,3).
Then i resorted to this technique, which, if you remember I had applied on a question posed by you earlier. This technique is tailor-made for probs like this, so it was pretty straighforward thereon
The link is http://targetiit.com/iit_jee_forum/posts/inequalities_3084.html
341\sum_{cyc} \frac{a}{2a+bc} = \sum_{cyc} \frac{a^2}{2a^2+abc} = \sum_{cyc} \frac{a^2}{2a^2+1} = \frac{1}{2} \left( 1 - \frac{1}{2a^2+1}\right) Hence the given inequality is true if \sum_{cyc} \frac{1}{2a^2+1} \ge 1
Now, we consider the function f(x) = \frac{1}{1+2e^x}
f"(x) = \frac{2e^x}{(1+2e^x)^2} \left[1-\frac{1}{1+2e^x} \right]>0
Let e^{x_1} = a^2; e^{x_2} = b^2; e^{x_3} = c^2
Hence, applying Jensen's inequality, we get,
\sum \frac{1}{1+2e^{x_1}} \ge \frac{3}{1+2e^{\frac{x_1+x_2+x_3}{3}}} = \frac{3}{1+ 2 (abc)^{\frac{2}{3}}}
Now, since, a+b+c = 3, abc ≤ 1.
Hence\frac{3}{1+ 2 (abc)^{\frac{2}{3}}} \ge 1
Thus, we have proved that \sum \frac{1}{2a^2+1} \ge 1 and hence the required inequality
39sir anyways we can do anything from where i stopped in #16 ? or os it totally hopeless?
1i mean we will try to minimise (3/(2+bc/a)), as its max will be 1.5 when min(abc) will attain its smallest value i.e 0, so i maximised (bc/a), but not checked by writing on paper...but i also did the same in my first post, cant we use that argument ??
39we need to prove:
\[ \sum_{cyclic}\frac{a}{2a+bc}\leq 1\rightarrow \sum_{cyclic}\frac{2a} {2a+bc}\leq 2 \]
\[ \rightarrow\sum_{cyclic}(1-\frac{2a}{2a+bc})\geq 1 \]
now how to proceed?
39yes sir if a≤bc we can also write bc≥ b (since a≥b) it means c≥b which contradicts the given condition.
1at the most a=b=c=1, put these values in the given expression we get 1/3+1/3+1/3 =1 so hence proved , any deviation in value of a,b,c from 1 will increase the vlaue of the given exp greater than 1. hence proved
341By Cauchy Schwarz inequality:
(x+y+z) \left(\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \right) \ge (a+b+c)^2
This is written more elegantly as
\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z} \right \ge \frac{(a+b+c)^2}{x+y+z}
1sir can u pls tell me how u wrote this step,
cyclic.(a2/(a2+2abc))≥(a+b+c)2/(a2+b2+c2+6abc)
3411st one by another method:
By Cauchy Schwarz (in Titu's Lemma form)
\sum_{cyc} \frac{a}{a+2bc} = \sum_{cyc} \frac{a^2}{a^2+2abc} \ge \frac{(a+b+c)^2}{a^2+b^2+c^2+6abc}
So now if we prove that(a+b+c)^2 \ge a^2+b^2+c^2+6abc we are done.
But, this is equivalent to proving 2\sum ab \ge 6abc \Rightarrow \sum \frac{1}{a} \ge 3
which is true as we have (a+b+c) \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \ge 9 and a+b+c = 3
1without\ loss\ of\ generality\ let\ a>b>c\\ \\ it\ implies\ \frac{ab}{c} >\frac{ac}{b}>\frac{bc}{a}\\ \\ or, \frac{a}{2a+bc}>\frac{b}{2b+ac}>\frac{c}{2c+ab}\\ \\ so,\frac{a}{2a+bc}+\frac{b}{2b+ac}+\frac{c}{2c+ab}\leq \frac{3}{(2+\frac{bc}{a})}\\ \\ now,to\ minimise\ RHS\ maximise\frac{bc}{a}\\ \\ from\ AM-GM\ max(\frac{bc}{a})=\frac{1}{max(abc)_{min}^2} =\frac{1}{a_{min}^2}=1,\\ stated\ in\ previous\ post.\\ \\ hence, \frac{a}{2a+bc}+\frac{b}{2b+ac}+\frac{c}{2c+ab}\leq1
1@ archna
by AM-GM u get
abc ≤ 1 and abc≤ (2a+bc)2/8
if x,y,z be some no's such that x≤y & x≤z , then we cannot conclude that z≥y or y≥z
to write 2a+bc ≥2√2 , u require minimum value of abc to be 1 , i.e abc ≥1 , which is not the case in this que. :(
1hope this is correct
2a+bc>2√2
as 2a+bc/2>√2abc & graetest value of abc=1
thrfor
a/2√2+b/2√2+c/2√2≤a/(2a+bc)+b/(2b+ca)+c/(2c+ab)≤1
3a/2√2<a/2√2+b/2√2+c/2√2≤a/(2a+bc)+b/(2b+ca)+c/(2c+ab)≤1
if a is least among 3 nos.
the greatest value of a<1
therfore 3a/2√2<1
39i have not gone through total but your first step itself is wrong. the inequality sign should have been opposite 
39ok , just an extension to the question
2] Let a,b,c be non-negative real numbers such that a+b+c=3. Prove that
\frac {a}{2a+bc}+\frac {b}{2b+ca}+\frac {c}{2c+ab}\le1
1by\ AM-GM \ inequality\ we\ get,\ abc\leq 1\\ \\ =>(\frac{a}{a+2bc}\geq \frac{a^2}{a^2+2})\ , (\frac{b}{b+2ac}\geq \frac{b^2}{b^2+2}),\ (\frac{c}{c+2ab}\geq \frac{c^2}{c^2+2})\\ \\ so\ adding\ all\ three\\ \\ \frac{a}{a+2bc}+\frac{b}{b+2ac}+\frac{c}{c+2ab}\geq \frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\\ \\ without\ the\ loss\ of\ generality\ let\ a>b>c\\ \\ we\ have\ \frac{b^2}{b^2+2}\geq \frac{b^2}{a^2+2},\ and\ \frac{c^2}{c^2+2}\geq \frac{c^2}{a^2+2}\\ \\ so,org\ inequation\ becomes\\ \\ \frac{a}{a+2bc}+\frac{b}{b+2ac}+\frac{c}{c+2ab}\geq \frac{a^2}{a^2+2}+\frac{b^2}{b^2+2}+\frac{c^2}{c^2+2}\geq \frac{a^2+b^2+c^2}{a^2+2}\\ \\ now,\ by\ power\ mean\ inequality\ a^2+b^2+c^2\geq 3\\ \\ so, \frac{a}{a+2bc}+\frac{b}{b+2ac}+\frac{c}{c+2ab}\geq \frac{3}{a^2+2}\\ \\ to\ maximise\ the\ RHS\ we\ have\ to\ minimise\ a \\ now\ minimum\ value\ of\ max(a,b,c)=1\\so\ a^2_{min}=1\\ \\ => \frac{a}{a+2bc}+\frac{b}{b+2ac}+\frac{c}{c+2ab}\geq 1
proved...
