Sides of a triangle

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

Yes it is a correct solution :)

We have

\frac{1}{y}+\frac{1}{z} \ge \frac{4}{y+z}

Hence

(x+y+z)\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \right) \ge (x+y+z)\left(\frac{1}{x}+\frac{4}{y+z} \right)

= 5+ \frac{3x}{y+z} + \frac{x}{y+z} + \frac{y+z}{x} \ge 7+\frac{3x}{y+z}

Now if x \ge y+z we get that the given expression is at least 10 which violates the condition.

Hence x<y+z and by symmetry, y<x+z, and z<x+y and we are done

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 :(

thank you very much

I went to INMO camp only if thats what youre saying ...

all the best to you too :) :)