1sir can we do this by lemma 's inequality...?
An ex-goiitian and present iitian is an orgainizer for an online math competition named Shaastra.
The first question is a nice inequality:
If x,y,z \in \mathbb{R^+} are such that x+y+z = 9(xyz)^{\frac{2}{3}} prove that
\frac{xy}{x+y+2z} + \frac{yz}{2x+y+z} + \frac{zx}{x+2y+z} \ge \frac{1}{4}
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8 Answers
1by lemma
xy/x+y+2z + yz/2x+y+z + zx/x+2y+z ≥ (xyz)^2/4(x+y+z)
what to do next...?
341Thats now how it goes arshad. Check up on how the lemma is used
Also a lemma is a shortened theorem. The lemma is named after a guy called Titu (Andreescu)
11(\frac{A_1}{a_1}+...\frac{A_n}{a_n})(a_1+a_2+...a_n)\ge n(A_1+A_2+...A_n)
If ai≥ai+1
So for that we choose the order of reals x,y,z as z≥x≥y
so reqd inequality becomes ≥ 3\frac{(xy+yz+zx)}{4(x+y+z)}
3(xy+yz+zx0\ge 9(\sqrt[3]{(xyz)^2} \\ \implies \frac{x+y+z}{3(xy+yz+zx)}\le 1\\
K=\frac{x+y+z}{3(xy+yz+zx)}\le 1
1k≥1
k4≥1
Thus proved!
111 edit - in the last step it is 14k≥14
and in #5 it is 3(xy+yz+xz)≥9.....
There's a zero in place of the bracket. [78]
341Soumik, that first inequality - are you sure about it. I dont think we conclusively proved it in the thread you posted.
But you are almost there. Good show! Just use a more conventional inequality and you will be through
341Precisely. So to make the solution lucid, all you needed to give was a statement on the ordering of the sequences and mentioned Chebyshev.