Inequality

sir please can you tell me where is the fault
setting a=\frac{x}{y},b=\frac{y}{z},c=\frac{z}{x}
so the inequality reduces to 17\ge 8(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})

from cauchy-schwarz, \left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)^2\le 3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right)

\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)^2\le 3\left[\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)^2-2(a+b+c) \right]
let \frac{1}{a}+\frac{1}{b}+\frac{1}{c} be m

so m^2 \le 3\left(m^2-4 \right)

m\ge\sqrt{6} or m\le\ - \sqrt{6}

ok. thanks nishant sir
i think a,b,c\ge0 as inequality is not valid for (a,b,c)=(-1,1,2)
CASE 1
assuming a=0 , b,c \ne0
8\ge 8(bc) is true by AM-GM on b,c
CASE 2
a,b=0 ,c \ne 0
8>0 which is true
CASE 3
a,b,c \ne 0
dividing the given inequality by abc we get
9+\frac{8}{abc} \ge 8\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right)
from titu's lemma, we have

\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \le \frac{9}{a+b+c}

or it is sufficient to show
9+ \frac{8}{abc} \ge 8.\frac{9}{2}

or abc \le \frac{8}{27}
which is true by AM-GM on a,b,c

8(ab+bc+ca)=16-4(a^2+b^2+c^2)
so ultimately I've to prove
4(a^2+b^2+c^2)+9abc\ge 8
Which is same as proving
(a+b+c)^2-(a+b+c)(a^2+b^2+c^2)-\frac{9}{2}abc\le 0

If i consider this to be a quadratic in (a+b+c), then i must have
\frac{a^2+b^2+c^2-\sqrt{(a^2+b^2+c^2+18abc)}}{2}<(a+b+c)<\frac{a^2+b^2+c^2+\sqrt{(a^2+b^2+c^2+18abc)}}{2}
So that means
(a²+b²+c²)-D≤4, which is true.....as (a²+b²+c²)=4-k, where k>0

for the 2nd inequality, we have by squarring and bring all terms to the L.H.S
[a^2+b^2+c^2+2(ab+bc+ca)](a^2+b^2+c^2-2[ab+bc+ca])+18abc>0
Which again is true, so proved that quadratic always lies below the X-axis.

not really. But knowing the Cauchy Schwarz inequality simplifies a host of problems. So if you have the time you can read it up. Titu's lemma is Cauchy Schwarz applied in a particular form that is elegant and useful.

Otherwise this is purely olympiad stuff as the sub-forum suggests.

A few days back prophet sir rightly said something - Sometimes the most obvious escapes us!

Bring all terms to the L.H.S
Then the expression can be easily factorised as a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)=k

Which is obviously non-negative by virtue of Schur's Inequality.

If the soln overlaps with the written lines, then to avert the confusion, i state that use 8=(a+b+c)³ for 8 in l.h.s
and 4(a+b+c) for 8 in r.h.s., now expand, and bring everything to l.h.s to get

9abc+(a^3+b^3+c^3)+3(a^2b+a^2c+ab^2+2abc+ac^2+b^c+bc^2)-(4a^2b+12abc+4ca^2+4ab^2+4[b^2c+bc^2+ac^2])\\ \ = a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)\ge 0