From Shaastra

sir can we do this by lemma 's inequality...?

by lemma
xy/x+y+2z + yz/2x+y+z + zx/x+2y+z ≥ (xyz)^2/4(x+y+z)
what to do next...?

Thats 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)

(\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 a_i≥a_i+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!

1 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]

Soumik, 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

but prophet sir, is that not simply chebychef on
X_is = A_ia_i
and Y_is=a_i

Precisely. So to make the solution lucid, all you needed to give was a statement on the ordering of the sequences and mentioned Chebyshev.