Sides of a triangle

Take x=1, y=2, z=3 then does it hold?

to show that x, y,z are sides of triangle it is enough to prove

x<y+z

so

(x+ y+ z ) * (1/x + 1/y + 1/z)

= 1 + 1+ 1+ x/y + x/y + y/x + y/z + z/x + z/y

= 10 > 3 + 2 + x( 1/y + 1/z) + 1/x * (y + z)

implies that

5> 2x/√yz + 2 √yz / x

let a = x/√yz

so

5 > 2a + 2a

solving this we get

1/2 < a < 2

so x/√zy < 2

x< 2√yz < y + z

Yes it is a correct solution :)

No I did not know the solution

Its actually IMO 2004

Here is my solution (morally all are the same ...)

Because the inequality is homogeneous, WLOG assume min{x,y,z} = z = 1

WLOG let x ≥y

so on the contrary it suffices to prove if x ≥ y+1 we have

\left ( x+y+1 \right )\left ( \frac{1}{x} + \frac{1}{y} + 1\right ) \ge 10

We have x \ge 1+ y \ge 2\sqrt{y} \implies x^2 \ge 4y (*)

After a short calculation we have to prove \frac xy + x + \frac yx + \frac 1x + y + \frac 1y \ge 7

Since y + \frac 1y \ge 2 it is enough to prove

\frac xy + x + \frac yx + \frac 1x \ge 5

Since x \ge y+1 \iff \frac xy \ge 1 + \frac 1y

It will be enough to prove \frac 1y + x + \frac yx + \frac 1x \ge 4

Which is true by AM-GM because \frac x4 + \frac x4 + \frac x4 + \frac x4 + \frac {1}{2y} + \frac {1}{2y} + \frac 1y + \frac yx \ge 8 \left ( \frac{x^2}{4^5 \cdot y} \right )^{1/8}

and by (*) 8 \left ( \frac{x^2}{4^5 \cdot y} \right )^{1/8} \ge 8 \cdot \frac 12 = 4

Actually it is the base case of IMO 2004, the original problem is easy induction after that

or even w/o induction it reduces to proving the same

Prophet sir gave a nice proof, which is not too much involved :)

No facebook likes so far :(

@ shubhodip :

how r u preparing for inmo ????

do you go to any classes???

i have not been able to prepare anything bcoz i have to prepare for boards.....so there r very less chances that i will be selected for inmo....

but all the best to u....