1cant we solve it by taking am>gm>hm??
16 Answers
1sir 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}
62the mistake is by taking a,b ,c the way you have,
you have taken that abc=1 which is not given!
1ok. 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
341You have reversed Titu's lemma. He is going to be very annoyed
1oh!! why cant i ever get a problem right using titu's lemma :((
118(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.
341not 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.
11Oops!
i just got carried away by thining A.m-G.M will do it.....It isn't!
I've to re-think it.
11A 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.
341yup thats the essence. Only you have to show how you got terms like a3 etc. to appear. there's an important step there. It will be nice if you take the time to give the complete solution
11Fine ,here's the complete soln.
8=(a+b+c)^3 - Use this for 8 in l.h.s.
For R.H.S, 8=4(a+b+c)
Now bring all terms to the L.H.S
Expansion gives - 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
..........
11If 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